JEE Binomial Theorem Questions

We need to find the coefficient of $$x^{30}$$ in the expansion of $$\left(1 + \frac{1}{x}\right)^6(1 + x^2)^7(1 - x^3)^8$$, where $$x \neq 0$$.

Rewrite the factor $$\left(1 + \frac{1}{x}\right)^6$$ as $$\frac{(1+x)^6}{x^6}$$ so that the expression becomes $$\frac{(1+x)^6(1+x^2)^7(1-x^3)^8}{x^6}$$ and hence the coefficient of $$x^{30}$$ in the original product equals the coefficient of $$x^{36}$$ in $$(1+x)^6(1+x^2)^7(1-x^3)^8$$.

By the binomial theorem, $$(1+x)^6=\sum_{a=0}^6\binom{6}{a}x^a\,,\quad(1+x^2)^7=\sum_{b=0}^7\binom{7}{b}x^{2b}\,,\quad(1-x^3)^8=\sum_{c=0}^8\binom{8}{c}(-1)^cx^{3c}\,,$$ so we require $$a+2b+3c=36$$ with $$0\le a\le6\,,\;0\le b\le7\,,\;0\le c\le8\,. $$

Since $$a+2b\le6+14=20$$, it follows that $$3c=36-(a+2b)\ge36-20=16$$ so $$c\ge6$$, while $$3c\le36$$ gives $$c\le12$$ and hence $$6\le c\le8$$.

Case c = 6: Then $$a+2b=36-18=18$$ and writing $$a=18-2b$$ with $$0\le a\le6$$ and $$0\le b\le7$$ shows that $$b\ge6$$ and $$b\le7$$, yielding $$(b,a)=(6,6)\text{ or }(7,4)\,. $$

Case c = 7: Then $$a+2b=36-21=15$$ and $$a=15-2b$$ with the same bounds gives $$b\ge5$$ and $$b\le7$$, so $$(b,a)=(5,5),(6,3),(7,1)\,. $$

Case c = 8: Then $$a+2b=36-24=12$$ and $$a=12-2b$$ yields $$b\ge3$$ and $$b\le6$$, hence $$(b,a)=(3,6),(4,4),(5,2),(6,0)\,. $$

The contribution of each term is $$\binom{6}{a}\binom{7}{b}\binom{8}{c}(-1)^c\,. $$ For c = 6 one has $$(-1)^6=1$$ and $$\binom{8}{6}=28$$, giving contributions $$\binom{6}{6}\binom{7}{6}\cdot28=196$$ and $$\binom{6}{4}\binom{7}{7}\cdot28=420\,. $$

For c = 7 one has $$(-1)^7=-1$$ and $$\binom{8}{7}=8$$, giving contributions $$\binom{6}{5}\binom{7}{5}\cdot8\cdot(-1)=-1008\,,\quad\binom{6}{3}\binom{7}{6}\cdot8\cdot(-1)=-1120\,,\quad\binom{6}{1}\binom{7}{7}\cdot8\cdot(-1)=-48\,. $$

For c = 8 one has $$(-1)^8=1$$ and $$\binom{8}{8}=1$$, giving contributions $$\binom{6}{6}\binom{7}{3}=35\,,\;\binom{6}{4}\binom{7}{4}=525\,,\;\binom{6}{2}\binom{7}{5}=315\,,\;\binom{6}{0}\binom{7}{6}=7\,. $$

Summing all contributions yields $$\alpha=196+420-1008-1120-48+35+525+315+7=-678\,$$ and a quick check shows the positive terms sum to 1498 while the negative terms sum to 2176, giving $$1498-2176=-678\,. $$

Therefore the absolute value is $$|\alpha|=678$$ and the answer is 678.

$$(a + bx + cx^2)^{10} = \sum p_i x^i$$

$$p_0 = a^{10}$$

For $$p_1$$: coefficient of $$x$$ comes from choosing $$bx$$ once and $$a$$ nine times:

$$p_1 = \binom{10}{1} a^9 \cdot b = 10a^9 b = 20$$

$$a^9 b = 2$$ ... (1)

Since $$a, b \in \mathbb{N}$$, the only solution to (1) is $$a = 1, b = 2$$.

For $$p_2$$: coefficient of $$x^2$$ comes from choosing $$bx$$ twice OR choosing $$cx^2$$ once:

$$p_2 = \binom{10}{2}a^8 b^2 + \binom{10}{1}a^9 c = 45 \times 1 \times 4 + 10 \times 1 \times c = 180 + 10c = 210$$

$$10c = 30$$, $$c = 3$$

$$2(a + b + c) = 2(1 + 2 + 3) = 12$$

This matches option 3: 12.

Given: Find the absolute difference of the coefficients of $$x^{10}$$ and $$x^7$$ in the expansion of $$\left(2x^2 + \frac{1}{2x}\right)^{11}$$.

General term of the expansion:

$$T_{r+1} = \binom{11}{r}(2x^2)^{11-r}\left(\frac{1}{2x}\right)^r = \binom{11}{r} \cdot 2^{11-r} \cdot x^{2(11-r)} \cdot 2^{-r} \cdot x^{-r}$$

$$= \binom{11}{r} \cdot 2^{11-2r} \cdot x^{22-3r}$$

Coefficient of $$x^{10}$$:

Set $$22 - 3r = 10$$, which gives $$r = 4$$.

$$\text{Coefficient} = \binom{11}{4} \cdot 2^{11-8} = 330 \times 2^3 = 330 \times 8 = 2640$$

Coefficient of $$x^7$$:

Set $$22 - 3r = 7$$, which gives $$r = 5$$.

$$\text{Coefficient} = \binom{11}{5} \cdot 2^{11-10} = 462 \times 2^1 = 462 \times 2 = 924$$

Absolute difference:

$$|2640 - 924| = 1716$$

Now, $$12^3 - 12 = 1728 - 12 = 1716$$ ✓

The correct answer is Option D: $$12^3 - 12$$.

We are given the binomial expansion of $$(\sqrt{2^{\log_2(10-3^x)}} + \sqrt[5]{2^{(x-2)\log_2 3}})^m$$.

First, simplify the terms. The first term is $$\sqrt{2^{\log_2(10-3^x)}} = (10-3^x)^{1/2}$$ and the second term is $$\sqrt[5]{2^{(x-2)\log_2 3}} = \bigl(2^{\log_2 3}\bigr)^{(x-2)/5} = 3^{(x-2)/5},$$ so the expression becomes $$( (10-3^x)^{1/2} + 3^{(x-2)/5} )^m$$.

Next, the binomial coefficients of the 2nd, 3rd, and 4th terms are $$\binom{m}{1}, \binom{m}{2}, \binom{m}{3},$$ which correspond to the 1st, 3rd, and 5th terms of an arithmetic progression. For these three terms to be equally spaced we require $$\binom{m}{2} - \binom{m}{1} = \binom{m}{3} - \binom{m}{2},$$ that is $$2\binom{m}{2} = \binom{m}{1} + \binom{m}{3}.$$ Substituting the values of the binomial coefficients gives $$m(m-1) = m + \frac{m(m-1)(m-2)}{6},$$ which simplifies to $$6(m-1) = 6 + (m-1)(m-2),$$ then $$6m - 6 = 6 + m^2 - 3m + 2,$$ leading to $$m^2 - 9m + 14 = 0$$ and hence $$(m-2)(m-7) = 0.$$ Since the 6th term must exist ($$m\ge5$$), we choose $$m=7$$.

Now using the 6th term condition, $$T_6 = \binom{7}{5} \cdot \bigl((10-3^x)^{1/2}\bigr)^2 \cdot \bigl(3^{(x-2)/5}\bigr)^5 = 21$$ implies $$21\,(10-3^x)\,3^{x-2} = 21$$ and thus $$(10-3^x)\,3^{x-2} = 1.$$ Letting $$t = 3^x$$ transforms this into $$(10 - t)\frac{t}{9} = 1,$$ which simplifies to $$10t - t^2 = 9,$$ then $$t^2 - 10t + 9 = 0,$$ so $$(t-1)(t-9) = 0.$$ Therefore $$t=1\implies x=0$$ or $$t=9\implies x=2.$$

Both values are valid since for $$x=0$$ we have $$10-1=9\gt 0$$ and for $$x=2$$ we have $$10-9=1\gt 0$$. The sum of squares of all possible values of $$x$$ is $$0^2 + 2^2 = 0 + 4 = 4$$.

We need to find the remainder when $$19^{200} + 23^{200}$$ is divided by 49.

Write $$19 = 21 - 2$$ and $$23 = 21 + 2$$. Then:

$$19^{200} + 23^{200} = (21-2)^{200} + (21+2)^{200}$$

Expanding using the binomial theorem, odd powers of 21 cancel, leaving:

$$= 2\sum_{k=0}^{100}\binom{200}{2k}(21)^{2k} \cdot 2^{200-2k}$$

Since $$21^2 = 441 = 9 \times 49 \equiv 0 \pmod{49}$$, all terms with $$k \geq 1$$ vanish modulo 49. Only the $$k = 0$$ term survives:

$$19^{200} + 23^{200} \equiv 2 \cdot 2^{200} = 2^{201} \pmod{49}$$

By Euler's theorem, $$\phi(49) = 42$$, so $$2^{42} \equiv 1 \pmod{49}$$.

$$201 = 42 \times 4 + 33$$, so $$2^{201} \equiv 2^{33} \pmod{49}$$.

Computing $$2^{33} \pmod{49}$$ step by step:

$$2^8 = 256 = 5 \times 49 + 11 \equiv 11$$

$$2^{16} \equiv 11^2 = 121 = 2 \times 49 + 23 \equiv 23$$

$$2^{32} \equiv 23^2 = 529 = 10 \times 49 + 39 \equiv 39$$

$$2^{33} = 2^{32} \cdot 2 \equiv 39 \times 2 = 78 = 49 + 29 \equiv 29$$

The remainder is 29.

We need to evaluate $$\displaystyle\sum_{\substack{i,j=0 \\ i \neq j}}^{n} \binom{n}{i}\binom{n}{j}$$.

$$\sum_{\substack{i,j=0 \\ i \neq j}}^{n} \binom{n}{i}\binom{n}{j} = \sum_{i=0}^{n}\sum_{j=0}^{n} \binom{n}{i}\binom{n}{j} - \sum_{i=0}^{n} \binom{n}{i}^2$$

$$\sum_{i=0}^{n}\sum_{j=0}^{n} \binom{n}{i}\binom{n}{j} = \left(\sum_{i=0}^{n}\binom{n}{i}\right)\left(\sum_{j=0}^{n}\binom{n}{j}\right) = 2^n \cdot 2^n = 2^{2n}$$

By Vandermonde's identity (or the Cauchy product):

$$\sum_{i=0}^{n} \binom{n}{i}^2 = \sum_{i=0}^{n} \binom{n}{i}\binom{n}{n-i} = \binom{2n}{n}$$

$$\sum_{\substack{i,j=0 \\ i \neq j}}^{n} \binom{n}{i}\binom{n}{j} = 2^{2n} - \binom{2n}{n}$$

Therefore, the correct answer is Option A: $$2^{2n} - \binom{2n}{n}$$.

Solution :

We need to find remainder when

$$7^{2022}+3^{2022}$$

is divided by $$5$$.

Now,

$$7 \equiv 2 \pmod 5$$

Therefore,

$$7^{2022} \equiv 2^{2022} \pmod 5$$

Also,

Powers of $$2$$ modulo $$5$$ repeat every 4 :

$$2^1 \equiv 2$$

$$2^2 \equiv 4$$

$$2^3 \equiv 3$$

$$2^4 \equiv 1 \pmod 5$$

Since,

$$2022 \equiv 2 \pmod 4$$

therefore,

$$2^{2022} \equiv 2^2 \equiv 4 \pmod 5$$

Now for $$3^{2022}$$ :

Powers of $$3$$ modulo $$5$$ also repeat every 4 :

$$3^1 \equiv 3$$

$$3^2 \equiv 4$$

$$3^3 \equiv 2$$

$$3^4 \equiv 1 \pmod 5$$

Again,

$$2022 \equiv 2 \pmod 4$$

Hence,

$$3^{2022} \equiv 3^2 \equiv 4 \pmod 5$$

Therefore,

$$7^{2022}+3^{2022} \equiv 4+4$$

$$\equiv 8$$

$$\equiv 3 \pmod 5$$

Final Answer :

$$3$$

We write the given expression as a sum: $$S = (5 + x)^{500} + x(5 + x)^{499} + x^2(5 + x)^{498} + \ldots + x^{500}$$ which can be expressed as $$S = \sum_{k=0}^{500} x^k (5 + x)^{500-k}\,.$$

Since this is a geometric series with first term $$(5+x)^{500}$$, common ratio $$\frac{x}{5+x}$$, and 501 terms, we have $$S = (5+x)^{500} \cdot \frac{1 - \left(\frac{x}{5+x}\right)^{501}}{1 - \frac{x}{5+x}}\,. $$ Substituting and simplifying gives $$= (5+x)^{500} \cdot \frac{\frac{(5+x)^{501} - x^{501}}{(5+x)^{501}}}{\frac{5}{5+x}} = \frac{(5+x)^{501} - x^{501}}{5}\,. $$

To find the coefficient of $$x^{101}$$, note that the term $$x^{501}$$ contributes zero since 101 < 501. The coefficient of $$x^{101}$$ in $$(5+x)^{501}$$ is $$\binom{501}{101} \cdot 5^{400}\,. $$ Dividing by 5 therefore yields $$\frac{\binom{501}{101} \cdot 5^{400}}{5} = \binom{501}{101} \cdot 5^{399}\,. $$ From the above, the answer is $$^{501}C_{101} \times 5^{399},$$ which is Option A.

We need the sum of coefficients of all positive even powers of $$x$$ in the expansion of $$\left(2x^3 + \frac{3}{x}\right)^{10}$$.

First, the general term of the expansion can be written as $$T_{r+1} = \binom{10}{r}(2x^3)^{10-r}\left(\frac{3}{x}\right)^r = \binom{10}{r} \cdot 2^{10-r} \cdot 3^r \cdot x^{30-4r}$$.

Since the power of $$x$$ is $$30 - 4r$$, which is always even, for it to be positive we require $$30 - 4r > 0 \implies r < 7.5 \implies r \leq 7$$. Thus for $$r = 0, 1, 2, \ldots, 7$$ the powers of $$x$$ are $$30, 26, 22, 18, 14, 10, 6, 2$$, all positive and even.

Hence the sum of the required coefficients is $$S = \sum_{r=0}^{7} \binom{10}{r} \cdot 2^{10-r} \cdot 3^r$$.

On the other hand, the total sum of all coefficients, found by setting $$x = 1$$, is $$\sum_{r=0}^{10} \binom{10}{r} \cdot 2^{10-r} \cdot 3^r = (2 + 3)^{10} = 5^{10}$$, so that $$S = 5^{10} - \sum_{r=8}^{10} \binom{10}{r} \cdot 2^{10-r} \cdot 3^r$$.

Computing the excluded terms gives:
$$r = 8: \binom{10}{8} \cdot 2^2 \cdot 3^8 = 45 \cdot 4 \cdot 6561 = 1180980$$
$$r = 9: \binom{10}{9} \cdot 2^1 \cdot 3^9 = 10 \cdot 2 \cdot 19683 = 393660$$
$$r = 10: \binom{10}{10} \cdot 2^0 \cdot 3^{10} = 59049$$
$$\text{Sum} = 1180980 + 393660 + 59049 = 1633689$$

Therefore
$$S = 5^{10} - 1633689 = 5^{10} - \beta \cdot 3^9$$
$$\beta = \frac{1633689}{3^9} = \frac{1633689}{19683} = 83$$
The answer is $$\boxed{83}$$.

We need to compute $$\displaystyle\sum_{r=0}^{5} \dfrac{1}{\binom{11}{r}}$$ and express it as $$\dfrac{a}{b}$$ in lowest terms, then find $$a - b$$.

The binomial coefficients are: $$\binom{11}{0} = 1$$, $$\binom{11}{1} = 11$$, $$\binom{11}{2} = 55$$, $$\binom{11}{3} = 165$$, $$\binom{11}{4} = 330$$, $$\binom{11}{5} = 462$$.

The sum is $$\dfrac{1}{1} + \dfrac{1}{11} + \dfrac{1}{55} + \dfrac{1}{165} + \dfrac{1}{330} + \dfrac{1}{462}$$.

Finding the LCM of the denominators: $$\text{lcm}(1, 11, 55, 165, 330, 462)$$. We have $$462 = 2 \cdot 3 \cdot 7 \cdot 11$$, $$330 = 2 \cdot 3 \cdot 5 \cdot 11$$, $$165 = 3 \cdot 5 \cdot 11$$, $$55 = 5 \cdot 11$$. So $$\text{lcm} = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310$$.

Converting each fraction: $$\dfrac{2310}{1} + \dfrac{2310}{11} + \dfrac{2310}{55} + \dfrac{2310}{165} + \dfrac{2310}{330} + \dfrac{2310}{462}$$

$$= 2310 + 210 + 42 + 14 + 7 + 5 = 2588$$

So the sum is $$\dfrac{2588}{2310}$$. Simplifying: $$\gcd(2588, 2310)$$. $$2588 = 2310 + 278$$, $$2310 = 8 \cdot 278 + 86$$, $$278 = 3 \cdot 86 + 20$$, $$86 = 4 \cdot 20 + 6$$, $$20 = 3 \cdot 6 + 2$$, $$6 = 3 \cdot 2$$. So $$\gcd = 2$$.

$$\dfrac{2588}{2310} = \dfrac{1294}{1155}$$

Checking if this is in lowest terms: $$1294 = 2 \cdot 647$$, $$1155 = 3 \cdot 5 \cdot 7 \cdot 11$$. Since $$647$$ is prime and does not divide $$1155$$, $$\gcd(1294, 1155) = 1$$.

Therefore $$a = 1294$$, $$b = 1155$$, and $$a - b = 1294 - 1155 = 139$$.

The answer is $$139$$.

Given,

$$C_r=\binom{10}{r}$$

Let

$$S=C_{1}+3\cdot2C_{2}+5\cdot3C_{3}+\cdots$$ upto $$10$$ terms

Now,

$$(1+x)^{10}=C_0+C_1x+C_2x^2+\cdots+C_{10}x^{10}$$

Differentiating,

$$10(1+x)^9=C_1+2C_2x+3C_3x^2+\cdots+10C_{10}x^9$$

Replace $$x$$ by $$x^2,$$

$$10(1+x^2)^9=C_1+2C_2x^2+3C_3x^4+\cdots+10C_{10}x^{18}$$

Multiplying by $$x,$$

$$10x(1+x^2)^9=C_1x+2C_2x^3+3C_3x^5+\cdots+10C_{10}x^{19}$$

Differentiating,

$$10\left((1+x^2)^9+x\cdot9(1+x^2)^8\cdot2x\right)=C_1+3\cdot2C_2x^2+5\cdot3C_3x^4+\cdots+19\cdot10C_{10}x^{18}$$

Putting $$x=1,$$

$$10\left(2^9+18\cdot2^8\right)=C_1+3\cdot2C_2+5\cdot3C_3+\cdots+19\cdot10C_{10}$$

Hence,

$$S=10\cdot2^8(2+18)$$

$$=10\cdot20\cdot2^8$$

$$=100\cdot2^9$$

Now,

$$T=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots+\frac{C_{10}}{11}$$

$$=\sum_{r=0}^{10}\frac{C_r}{r+1}$$

Since,

$$C_r=\binom{10}{r}$$

we use the identity

$$\frac1{r+1}\binom{10}{r}=\frac1{11}\binom{11}{r+1}$$

Therefore,

$$T=\frac1{11}\sum_{r=0}^{10}\binom{11}{r+1}$$

Now put

$$k=r+1$$

Then,

$$T=\frac1{11}\sum_{k=1}^{11}\binom{11}{k}$$

Using,

$$\sum_{k=0}^{11}\binom{11}{k}=2^{11}$$

we get

$$\sum_{k=1}^{11}\binom{11}{k}=2^{11}-1$$

Hence,

$$T=\frac{2^{11}-1}{11}$$

Previously,

$$S=C_1+3\cdot2C_2+5\cdot3C_3+\cdots+19\cdot10C_{10}=100\cdot2^9$$

Given,

$$S=\frac{\alpha\cdot2^{11}}{2^\beta-1}T$$

Substituting the values of $$S$$ and $$T,$$

$$100\cdot2^9=\frac{\alpha\cdot2^{11}}{2^\beta-1}\cdot\frac{2^{11}-1}{11}$$

Comparing denominator terms,

$$2^\beta-1=2^{11}-1$$

Hence,

$$\beta=11$$

Now,

$$100\cdot2^9=\frac{\alpha\cdot2^{11}}{11}$$

Dividing by $$2^9,$$

$$100=\frac{4\alpha}{11}$$

$$1100=4\alpha$$

$$\alpha=275$$

Therefore,

$$\alpha+\beta=275+11=286$$

Hence, $$\boxed{286}$$.

We are given $$(1 - x + x^3)^n = \sum_{j=0}^{3n} a_j x^j$$ and need to find $$\sum_{j=0}^{[3n/2]} a_{2j} + 4\sum_{j=0}^{[(3n-1)/2]} a_{2j+1}$$.

Substituting $$x = 1$$: $$(1 - 1 + 1)^n = 1 = \sum_{j=0}^{3n} a_j$$. This gives $$\sum a_{2j} + \sum a_{2j+1} = 1$$.

Substituting $$x = -1$$: $$(1 + 1 - 1)^n = 1 = \sum_{j=0}^{3n} a_j(-1)^j$$. This gives $$\sum a_{2j} - \sum a_{2j+1} = 1$$.

Adding these two equations: $$2\sum a_{2j} = 2$$, so $$\sum a_{2j} = 1$$. Subtracting: $$2\sum a_{2j+1} = 0$$, so $$\sum a_{2j+1} = 0$$.

Therefore, the required expression equals $$\sum a_{2j} + 4\sum a_{2j+1} = 1 + 4(0) = 1$$.

We need the coefficient of $$x^{256}$$ in $$(1-x)^{101}(x^2+x+1)^{100}$$.

Observe the factorisation $$1 - x^3 = (1-x)(x^2+x+1)$$, so $$(x^2+x+1)^{100} = \frac{(1-x^3)^{100}}{(1-x)^{100}}$$.

Therefore the expression becomes: $$(1-x)^{101} \cdot \frac{(1-x^3)^{100}}{(1-x)^{100}} = (1-x)(1-x^3)^{100}.$$

We need the coefficient of $$x^{256}$$ in $$(1-x)(1-x^3)^{100}$$, which equals $$[\text{coeff of }x^{256}\text{ in }(1-x^3)^{100}] - [\text{coeff of }x^{255}\text{ in }(1-x^3)^{100}].$$

The general term in $$(1-x^3)^{100}$$ is $$\binom{100}{k}(-1)^k x^{3k}$$, so only powers that are multiples of 3 appear.

Since $$256$$ is not a multiple of $$3$$, the coefficient of $$x^{256}$$ in $$(1-x^3)^{100}$$ is $$0$$.

Since $$255 = 3 \times 85$$, the coefficient of $$x^{255}$$ in $$(1-x^3)^{100}$$ is $$\binom{100}{85}(-1)^{85} = -\binom{100}{85} = -\binom{100}{15}$$.

Therefore the coefficient of $$x^{256}$$ in the original expression is $$0 - \left(-\binom{100}{15}\right) = \binom{100}{15}.$$

We begin with the well-known Binomial Theorem, which states that for any non-negative integer $$n$$,

$$ (1+x)^{\,n}= \sum_{k=0}^{n} {^{\,n}C_k}\,x^{\,k}. $$

Now, in our question we see the expression $$\displaystyle\sum_{k=0}^{20}\bigl({^{20}C_k}\bigr)^2$$. A classical way to evaluate such a square-of-coefficients sum is to multiply two identical binomial expansions and compare coefficients.

First, write down two identical expansions of $$(1+x)^{20}$$:

$$ (1+x)^{20}= \sum_{k=0}^{20}{^{20}C_k}\,x^{\,k}, \qquad (1+x)^{20}= \sum_{r=0}^{20}{^{20}C_r}\,x^{\,r}. $$

Multiplying these two series term by term we get

$$ (1+x)^{20}\,(1+x)^{20}= (1+x)^{40}. $$

On the left-hand side, the Cauchy product of the two sums is

$$ \Bigl(\sum_{k=0}^{20}{^{20}C_k}\,x^{\,k}\Bigr) \Bigl(\sum_{r=0}^{20}{^{20}C_r}\,x^{\,r}\Bigr) = \sum_{m=0}^{40}\;\Bigl(\sum_{k=0}^{m}{^{20}C_k}\,{^{20}C_{m-k}}\Bigr)\,x^{\,m}. $$

In particular, we are interested in the coefficient of $$x^{20}$$. Setting $$m=20$$ in the inner sum we have

$$ \sum_{k=0}^{20}{^{20}C_k}\,{^{20}C_{20-k}}. $$

But the symmetry property of binomial coefficients gives $$ {^{20}C_{20-k}}={^{20}C_k}. $$ Therefore the coefficient of $$x^{20}$$ on the left becomes

$$ \sum_{k=0}^{20}\bigl({^{20}C_k}\bigr)^2. $$

On the right-hand side, the expansion of $$(1+x)^{40}$$ using the Binomial Theorem is

$$ (1+x)^{40}= \sum_{m=0}^{40}{^{40}C_m}\,x^{\,m}. $$

Hence the coefficient of $$x^{20}$$ on the right is simply $$ {^{40}C_{20}}. $$

Since both series represent the same polynomial $$(1+x)^{40}$$, their coefficients of the same powers of $$x$$ must be equal. Setting the two expressions for the coefficient of $$x^{20}$$ equal, we get

$$ \sum_{k=0}^{20}\bigl({^{20}C_k}\bigr)^2 \;=\; {^{40}C_{20}}. $$

Thus, the required sum equals $$ {^{40}C_{20}} $$. This matches Option C.

Hence, the correct answer is Option C.

We are given that when $$x$$ is divided by 4, the remainder is 3, so $$x = 4k + 3$$ for some non-negative integer $$k$$.

We need the remainder when $$(2020 + x)^{2022}$$ is divided by 8. Substituting $$x = 4k + 3$$, we get $$2020 + x = 2020 + 4k + 3 = 2023 + 4k$$.

Now $$2023 + 4k \pmod{8}$$: since $$2023 = 252 \times 8 + 7$$, we have $$2023 \equiv 7 \pmod{8}$$, so $$2023 + 4k \equiv 7 + 4k \pmod{8}$$. Depending on whether $$k$$ is even or odd, this is either $$7$$ or $$3 \pmod{8}$$. In both cases, the number is odd.

Let $$m = 2023 + 4k$$. Since $$m$$ is odd, we can write $$m = 2q + 1$$. Then $$m^2 = 4q^2 + 4q + 1 = 4q(q+1) + 1$$. Since one of $$q, q+1$$ is even, $$q(q+1)$$ is even, so $$4q(q+1) \equiv 0 \pmod{8}$$. Thus $$m^2 \equiv 1 \pmod{8}$$.

Therefore $$m^{2022} = (m^2)^{1011} \equiv 1^{1011} = 1 \pmod{8}$$.

The remainder when $$(2020 + x)^{2022}$$ is divided by 8 is $$1$$.

We have to find the coefficient of the term containing $$a^{7} b^{8}$$ in the expansion of $$(a + 2b + 4ab)^{10}$$.

First recall the multinomial theorem. For any three terms $$x_{1}, x_{2}, x_{3}$$,

$$ (x_{1} + x_{2} + x_{3})^{10} = \sum_{n_{1}+n_{2}+n_{3}=10} \dfrac{10!}{n_{1}! \, n_{2}! \, n_{3}!}\; x_{1}^{\,n_{1}} x_{2}^{\,n_{2}} x_{3}^{\,n_{3}}. $$

In our question we identify

$$x_{1}=a, \qquad x_{2}=2b, \qquad x_{3}=4ab.$$

Let $$n_{1}, n_{2}, n_{3}$$ be the number of times these three terms are chosen, so

$$n_{1}+n_{2}+n_{3}=10.$$

The powers of $$a$$ and $$b$$ contributed by one choice of each term are

$$x_{1}=a \;\; \longrightarrow \;\; a^{1} b^{0},$$ $$x_{2}=2b \;\; \longrightarrow \;\; a^{0} b^{1},$$ $$x_{3}=4ab \;\;\longrightarrow \;\; a^{1} b^{1}.$

Hence after making $$n_{1}, n_{2}, n_{3}$$ selections, the total exponents of $$a$$ and $$b$$ become

$$\text{Power of } a : \; n_{1} + n_{3},$$ $$\text{Power of } b : \; n_{2} + n_{3}.$$

We need these powers to be $$7$$ and $$8$$ respectively, so we write the two equations

$$n_{1} + n_{3} = 7, \qquad n_{2} + n_{3} = 8.$$

Together with $$n_{1}+n_{2}+n_{3}=10,$$ we now solve for $$n_{1}, n_{2}, n_{3}.$

Adding the first two equations gives

$$n_{1}+n_{2}+2n_{3}=15.$$

But $$n_{1}+n_{2}=10-n_{3},$$ so substituting,

$$(10-n_{3}) + 2n_{3} = 15 \;\; \Longrightarrow \;\; 10 + n_{3} = 15 \;\; \Longrightarrow \;\; n_{3} = 5.$$

Now we back-substitute:

$$n_{1} = 7 - n_{3} = 7 - 5 = 2,$$ $$n_{2} = 8 - n_{3} = 8 - 5 = 3.$$

Thus the required term arises from $$n_{1}=2,\, n_{2}=3,\, n_{3}=5.$

According to the multinomial theorem, the coefficient of this term is

$$\dfrac{10!}{n_{1}! \, n_{2}! \, n_{3}!}\; (2)^{n_{2}}\;(4)^{n_{3}}.$$

Writing out the factorials,

$$10! = 3628800,$$ $$n_{1}! = 2! = 2,$$ $$n_{2}! = 3! = 6,$$ $$n_{3}! = 5! = 120.$$

So

$$\dfrac{10!}{2!\,3!\,5!} = \dfrac{3628800}{2 \times 6 \times 120} = \dfrac{3628800}{1440} = 2520.$$

Next incorporate the numerical factors from the bases:

$$(2)^{n_{2}} = 2^{3} = 8,$$ $$(4)^{n_{3}} = (2^{2})^{5} = 2^{10} = 1024.$$

Multiplying these powers of $$2$$,

$$8 \times 1024 = 2^{3} \times 2^{10} = 2^{13} = 8192.$$

Hence the full numerical coefficient of the term $$a^{7} b^{8}$$ is

$$2520 \times 8192 = 20643840.$$

We are told in the question that this coefficient can be written as $$K \cdot 2^{16}.$$ Since $$2^{16}=65536,$$ we set up the equation

$$K \times 65536 = 20643840.$$

Dividing to isolate $$K,$$

$$K = \dfrac{20643840}{65536}.$$

Compute the division step by step:

$$65536 \times 300 = 19660800,$$

Subtracting, $$20643840 - 19660800 = 983040,$$

$$65536 \times 10 = 655360,$$

Subtracting, $$983040 - 655360 = 327680,$$

$$65536 \times 5 = 327680,$$

Subtracting, the remainder is $$0.$$

Adding the multipliers, $$300 + 10 + 5 = 315.$$ Therefore,

$$K = 315.$$

So, the answer is $$315$$.

The general term in the expansion of $$\left(2x^r + \frac{1}{x^2}\right)^{10}$$ is given by

$$T_{k+1} = \binom{10}{k}\left(2x^r\right)^{10-k}\left(\frac{1}{x^2}\right)^k = \binom{10}{k} \cdot 2^{10-k} \cdot x^{r(10-k) - 2k}$$

For the constant term, the power of $$x$$ must be zero, so we need $$r(10 - k) - 2k = 0$$.

Solving for $$k$$: $$10r - rk - 2k = 0$$, which gives $$k(r + 2) = 10r$$, so $$k = \frac{10r}{r + 2}$$. $$-(1)$$

Since $$k$$ must be a non-negative integer with $$0 \le k \le 10$$, and the constant term equals 180, we also need $$\binom{10}{k} \cdot 2^{10-k} = 180$$. $$-(2)$$

We test values of $$k$$ in equation $$(2)$$:

For $$k = 8$$: $$\binom{10}{8} \cdot 2^{2} = 45 \times 4 = 180$$. This works.

Substituting $$k = 8$$ into equation $$(1)$$: $$8 = \frac{10r}{r + 2}$$.

Cross-multiplying: $$8(r + 2) = 10r$$, so $$8r + 16 = 10r$$, giving $$2r = 16$$, hence $$r = 8$$.

Therefore, the value of $$r$$ is $$8$$.

We begin with the standard binomial expansion

$$ (x+y)^n \;=\; \sum_{k=0}^{n} \binom{n}{k} \, x^{\,n-k}\,y^{\,k}. $$

The “sum of the coefficients’’ is obtained by putting $$x=1$$ and $$y=1$$, because each term $$\binom{n}{k}\,x^{\,n-k}y^{\,k}$$ then reduces to just $$\binom{n}{k}$$. Thus we have

$$ (1+1)^n \;=\; 2^{\,n} \;=\; \sum_{k=0}^{n} \binom{n}{k}. $$

The question tells us that this sum equals $$4096$$, so

$$ 2^{\,n} \;=\; 4096. $$

We now solve for $$n$$. Recognising $$4096$$ as a power of $$2$$, we write

$$ 4096 = 2^{12}, $$

and therefore

$$ n = 12. $$

So the expansion we are really interested in is

$$ (x+y)^{12}. $$

Next, we must find the greatest (i.e. the largest numerical) binomial coefficient among $$\binom{12}{0},\binom{12}{1},\dots,\binom{12}{12}$$. A basic result from the theory of binomial coefficients states:

The sequence $$\binom{n}{0},\binom{n}{1},\dots,\binom{n}{n}$$ increases up to $$k=\left\lfloor\dfrac{n}{2}\right\rfloor$$ and then decreases symmetrically.

Since $$n=12$$ is even, the maximum value occurs at

$$ k = \dfrac{n}{2} = \dfrac{12}{2} = 6. $$

The corresponding coefficient is therefore

$$ \binom{12}{6}. $$

We compute it explicitly using the formula $$\displaystyle \binom{n}{k} = \dfrac{n!}{k!\,(n-k)!}$$:

$$ \binom{12}{6} = \frac{12!}{6!\,6!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6!\, (6 \times 5 \times 4 \times 3 \times 2 \times 1)} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1}. $$

We simplify step by step:

First cancel the factor $$6$$:

$$ \frac{12}{6} = 2 $$

so one pair of factors becomes $$2$$. Proceeding orderly, we have

$$ \frac{2 \times 11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1}. $$

Now cancel a factor $$5$$ with the $$10$$:

$$ 10 \div 5 = 2, $$

giving

$$ \frac{2 \times 11 \times 2 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}. $$

Next cancel a factor $$4$$ with the $$8$$:

$$ 8 \div 4 = 2, $$

so we obtain

$$ \frac{2 \times 11 \times 2 \times 9 \times 2 \times 7}{3 \times 2 \times 1}. $$

Cancel a factor $$3$$ with the $$9$$:

$$ 9 \div 3 = 3, $$

giving

$$ \frac{2 \times 11 \times 2 \times 3 \times 2 \times 7}{2 \times 1}. $$

Finally cancel the remaining denominator $$2$$ with one of the numerators $$2$$, leaving

$$ 11 \times 2 \times 3 \times 2 \times 7. $$

Multiplying in order,

$$ 11 \times 2 = 22, \\ 22 \times 3 = 66, \\ 66 \times 2 = 132, \\ 132 \times 7 = 924. $$

Thus

$$ \binom{12}{6} = 924. $$

Hence, the correct answer is Option 924.

We have $$A = \sum_{k=0}^{n} (-1)^k \binom{n}{k}\left[\left(\frac{1}{2}\right)^k + \left(\frac{3}{4}\right)^k + \left(\frac{7}{8}\right)^k + \left(\frac{15}{16}\right)^k + \left(\frac{31}{32}\right)^k\right]$$.

This can be split into five sums, each of the form $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}r^k = (1-r)^n$$ by the binomial theorem.

So $$A = \left(1-\frac{1}{2}\right)^n + \left(1-\frac{3}{4}\right)^n + \left(1-\frac{7}{8}\right)^n + \left(1-\frac{15}{16}\right)^n + \left(1-\frac{31}{32}\right)^n$$.

$$A = \frac{1}{2^n} + \frac{1}{4^n} + \frac{1}{8^n} + \frac{1}{16^n} + \frac{1}{32^n} = \frac{1}{2^n} + \frac{1}{2^{2n}} + \frac{1}{2^{3n}} + \frac{1}{2^{4n}} + \frac{1}{2^{5n}}$$.

This is a geometric series with first term $$\frac{1}{2^n}$$ and ratio $$\frac{1}{2^n}$$, so $$A = \frac{\frac{1}{2^n}(1 - \frac{1}{2^{5n}})}{1 - \frac{1}{2^n}} = \frac{1 - \frac{1}{2^{5n}}}{2^n - 1}$$.

Now $$63A = 1 - \frac{1}{2^{30}}$$ gives us $$63 \cdot \frac{1 - \frac{1}{2^{5n}}}{2^n - 1} = 1 - \frac{1}{2^{30}}$$.

Trying $$n = 6$$: $$2^n - 1 = 63$$ and $$5n = 30$$. Then $$63 \cdot \frac{1 - \frac{1}{2^{30}}}{63} = 1 - \frac{1}{2^{30}}$$, which is satisfied.

The answer is $$n = 6$$.

We begin with the given series

$$^nC_0,\; 3\cdot {}^nC_1,\; 5\cdot {}^nC_2,\; 7\cdot {}^nC_3,\; \ldots,\;(n+1)\text{ terms}$$

Every coefficient in front of the binomial term is an odd number. In fact the $$k^{\text{th}}$$ term (starting with $$k=0$$) is

$$ (2k+1)\;{}^nC_k. $$

Therefore the required sum can be written compactly as

$$ S \;=\; \sum_{k=0}^{n} (2k+1)\;{}^nC_k. $$

We are told that this sum equals $$2^{100}\cdot 101$$, i.e.

$$ \sum_{k=0}^{n} (2k+1)\;{}^nC_k \;=\; 2^{100}\cdot 101. \quad -(1) $$

Now let us evaluate the left‐hand side purely algebraically. We separate the part involving $$k$$ and the part not involving $$k$$:

$$ \sum_{k=0}^{n} (2k+1)\;{}^nC_k \;=\; 2\sum_{k=0}^{n} k\;{}^nC_k \;+\; \sum_{k=0}^{n} {}^nC_k. $$

For the two standard binomial sums that appear here, we first state the well-known identities:

• Binomial sum $$\displaystyle\sum_{k=0}^{n} {}^nC_k = 2^n.$$
• Weighted binomial sum $$\displaystyle\sum_{k=0}^{n} k\;{}^nC_k = n\,2^{\,n-1}.$$

Using these, we obtain

$$ \begin{aligned} \sum_{k=0}^{n} (2k+1)\;{}^nC_k &= 2\Bigl(n\,2^{\,n-1}\Bigr) \;+\; 2^n \\ &= n\,2^{\,n} \;+\; 2^{\,n} \\ &= (n+1)\,2^{\,n}. \\ \end{aligned} $$

Hence the equation (1) becomes

$$ (n+1)\,2^{\,n} \;=\; 2^{100}\cdot 101. \quad -(2) $$

To compare the two sides, we isolate the odd and even (power-of-two) parts. Every positive integer can be written uniquely as $$\text{(odd part)}\times 2^{\text{(power of 2 present)}}.$$ Let us denote by $$v_2(m)$$ the exponent of the highest power of $$2$$ dividing $$m$$.

Write $$n+1 = 2^{\,r}\,m,$$ where $$m$$ is odd and $$r = v_2(n+1).$$ Then the left-hand side of (2) has a total power of two equal to $$n + r,$$ and its odd part equals $$m.$$ Comparing with the right-hand side, whose power of two is $$100$$ and whose odd part is $$101,$$ we obtain two simultaneous requirements:

$$ \begin{cases} m \;=\; 101,\\[4pt] n + r \;=\; 100. \end{cases} $$

But since $$m=101,$$ we have

$$ n+1 = 101\;2^{\,r}. \quad -(3) $$

Substituting $$n = 101\;2^{\,r} - 1$$ into the second requirement $$n + r = 100,$$ we get

$$ 101\;2^{\,r} - 1 + r \;=\; 100 \;\;\Longrightarrow\;\; 101\;2^{\,r} + r \;=\; 101. \quad -(4) $$

Because $$2^{\,r} \ge 1$$ and $$r \ge 0,$$ the left-hand side of (4) increases rapidly with $$r.$$ Testing non-negative integers:

• If $$r = 0$$ then $$101\cdot 1 + 0 = 101,$$ which satisfies (4).
• If $$r \ge 1$$ then $$101\;2^{\,r} \ge 202,$$ so the left-hand side exceeds $$101,$$ violating (4).

Thus $$r = 0$$ is the only admissible value. Feeding this back into (3) yields

$$ n+1 = 101\cdot 2^{\,0} = 101 \;\;\Longrightarrow\;\; n = 100. $$

The quantity asked for in the problem is

$$ 2\left[\frac{n-1}{2}\right]. $$

With $$n = 100,$$ we have

$$ \frac{n-1}{2} = \frac{100 - 1}{2} = \frac{99}{2} = 49.5. $$

The greatest integer function gives $$\left[\frac{99}{2}\right] = 49.$$

Multiplying by $$2$$ we obtain

$$ 2 \times 49 = 98. $$

So, the answer is $$98$$.

Using the Vandermonde identity, the first sum $$\sum_{i=0}^{k} \binom{10}{i}\binom{15}{k-i} = \binom{25}{k}$$, since $$\binom{n}{r} = 0$$ when $$r > n$$ or $$r < 0$$. Similarly, the second sum $$\sum_{i=0}^{k+1} \binom{12}{i}\binom{13}{k+1-i} = \binom{25}{k+1}$$.

So we need to find the maximum value of $$k$$ for which $$\binom{25}{k} + \binom{25}{k+1}$$ is maximized.

By Pascal's identity, $$\binom{25}{k} + \binom{25}{k+1} = \binom{26}{k+1}$$. The binomial coefficient $$\binom{26}{m}$$ is maximized when $$m = 13$$ (since 26 is even, the maximum occurs uniquely at $$m = 13$$). Setting $$k + 1 = 13$$ gives $$k = 12$$.

Therefore, the maximum value of $$k$$ is $$12$$.

The correct answer is $$12$$.

We have been given the notation

$$\left[\frac{n}{k}\right]=\begin{cases} \displaystyle\binom{n}{k}, & 0\le k\le n\\[4pt] 0, & \text{otherwise} \end{cases}$$

and the expression

$$A_k=\sum_{i=0}^{9}\binom{9}{i}\left[\frac{12}{\,12-k+i\,}\right]+\sum_{i=0}^{8}\binom{8}{i}\left[\frac{13}{\,13-k+i\,}\right].$$

Only those terms survive in which the lower argument of the bracket lies between $$0$$ and the upper argument inclusive. Thus, for the first summation the condition is

$$0\le 12-k+i\le 12\quad\Longrightarrow\quad k-12\le i\le k,$$

and for the second summation the condition is

$$0\le 13-k+i\le 13\quad\Longrightarrow\quad k-13\le i\le k.$$

Since $$i$$ is non-negative in both sums, the effective range is always $$0\le i\le k$$ (because $$k\le 13$$ in our problem). We shall now evaluate $$A_3$$ and $$A_4$$ separately.

1. Evaluation of $$A_3$$

When $$k=3$$ the admissible indices are $$i=0,1,2,3$$.

First sum — put $$k=3$$:

$$\left[\frac{12}{\,12-3+i\,}\right]=\binom{12}{\,9+i\,},$$ so $$\sum_{i=0}^{3}\binom{9}{i}\binom{12}{\,9+i\,}$$

Term-wise calculation:

$$\begin{aligned} i=0:&\; \binom{9}{0}\binom{12}{9}=1\cdot220=220\\ i=1:&\; \binom{9}{1}\binom{12}{10}=9\cdot66=594\\ i=2:&\; \binom{9}{2}\binom{12}{11}=36\cdot12=432\\ i=3:&\; \binom{9}{3}\binom{12}{12}=84\cdot 1=84 \end{aligned}$$

Adding,

$$220+594+432+84=1330.$$

Second sum — put $$k=3$$:

$$\left[\frac{13}{\,13-3+i\,}\right]=\binom{13}{\,10+i\,},$$ so $$\sum_{i=0}^{3}\binom{8}{i}\binom{13}{\,10+i\,}$$

Term-wise calculation:

$$\begin{aligned} i=0:&\; \binom{8}{0}\binom{13}{10}=1\cdot286=286\\ i=1:&\; \binom{8}{1}\binom{13}{11}=8\cdot78=624\\ i=2:&\; \binom{8}{2}\binom{13}{12}=28\cdot13=364\\ i=3:&\; \binom{8}{3}\binom{13}{13}=56\cdot 1=56 \end{aligned}$$

Adding,

$$286+624+364+56=1330.$$

Therefore

$$A_3=1330+1330=2660.$$

2. Evaluation of $$A_4$$

When $$k=4$$ the admissible indices are $$i=0,1,2,3,4$$.

First sum — put $$k=4$$:

$$\left[\frac{12}{\,12-4+i\,}\right]=\binom{12}{\,8+i\,},$$ so $$\sum_{i=0}^{4}\binom{9}{i}\binom{12}{\,8+i\,}$$

Term-wise calculation:

$$\begin{aligned} i=0:&\; \binom{9}{0}\binom{12}{8}=1\cdot495=495\\ i=1:&\; \binom{9}{1}\binom{12}{9}=9\cdot220=1980\\ i=2:&\; \binom{9}{2}\binom{12}{10}=36\cdot66=2376\\ i=3:&\; \binom{9}{3}\binom{12}{11}=84\cdot12=1008\\ i=4:&\; \binom{9}{4}\binom{12}{12}=126\cdot 1=126 \end{aligned}$$

Adding,

$$495+1980+2376+1008+126=5985.$$

Second sum — put $$k=4$$:

$$\left[\frac{13}{\,13-4+i\,}\right]=\binom{13}{\,9+i\,},$$ so $$\sum_{i=0}^{4}\binom{8}{i}\binom{13}{\,9+i\,}$$

Term-wise calculation:

$$\begin{aligned} i=0:&\; \binom{8}{0}\binom{13}{9}=1\cdot715=715\\ i=1:&\; \binom{8}{1}\binom{13}{10}=8\cdot286=2288\\ i=2:&\; \binom{8}{2}\binom{13}{11}=28\cdot78=2184\\ i=3:&\; \binom{8}{3}\binom{13}{12}=56\cdot13=728\\ i=4:&\; \binom{8}{4}\binom{13}{13}=70\cdot 1=70 \end{aligned}$$

Adding,

$$715+2288+2184+728+70=5985.$$

Therefore

$$A_4=5985+5985=11970.$$

3. Required difference

We now compute

$$A_4-A_3=11970-2660=9310.$$

The question states that

$$A_4-A_3=190\,p,$$ so $$p=\frac{9310}{190}=49.$$

Hence, the correct answer is Option 49.

For every binomial coefficient we recall the symmetry and monotonicity property

$$^nC_r = \; ^nC_{n-r}$$

and, starting from $$r = 0$$, the values of $$^nC_r$$ keep on increasing as $$r$$ increases up to the integral part of $$\dfrac{n}{2}$$, after which they start decreasing. This means that the greatest term (or terms) of the expansion of $$(1+x)^n$$ occur(s) at

$$r = \left\lfloor \dfrac{n}{2} \right\rfloor \quad\text{or}\quad r = \left\lceil \dfrac{n}{2} \right\rceil.$$

Now we apply this fact one by one.

For $$^{19}C_p$$ we have $$n = 19$$. The integral part of $$\dfrac{19}{2}$$ is $$9$$ and the next integer is $$10$$. Hence the greatest value is attained at $$p = 9$$ or $$p = 10$$, and both values are the same:

$$a = \;^{19}C_9 = \;^{19}C_{10}.$$

Using the factorial definition,

$$a = ^{19}C_9 = \dfrac{19!}{9!\,10!}.$$

Simplifying step by step,

$$\begin{aligned} a &= \dfrac{19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11}{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1} \\ &= 92378. \end{aligned}$$

For $$^{20}C_q$$ we have $$n = 20$$. Now $$\dfrac{20}{2}=10$$, so the single greatest value is at $$q = 10$$.

$$b = \;^{20}C_{10} = \dfrac{20!}{10!\,10!}.$$

Again expanding only what is needed,

$$\begin{aligned} b &= \dfrac{20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1} \\ &= 184756. \end{aligned}$$

For $$^{21}C_r$$ we take $$n = 21$$. Now $$\dfrac{21}{2}=10.5$$, so the two central positions $$r = 10$$ and $$r = 11$$ give the maximum and are equal:

$$c = \;^{21}C_{10} = \;^{21}C_{11}.$$

Using the recursive relation $$^{n}C_{r} = \;^{\,n-1}C_{r} + \;^{\,n-1}C_{r-1},$$

$$\begin{aligned} c &= ^{21}C_{10} \\ &= ^{20}C_{10} + \;^{20}C_{9} \\ &= 184756 + 167960 \\ &= 352716. \end{aligned}$$

We have therefore obtained

$$a = 92378,\qquad b = 184756,\qquad c = 352716.$$

Now we examine the ratios suggested by the four options. Start by dividing each number by the respective denominators in Option C:

$$ \begin{aligned} \dfrac{a}{11} &= \dfrac{92378}{11} = 8398,\\[4pt] \dfrac{b}{22} &= \dfrac{184756}{22} = 8398,\\[4pt] \dfrac{c}{42} &= \dfrac{352716}{42} = 8398. \end{aligned} $$

All three quotients are exactly equal to $$8398$$, so the three fractions are equal:

$$\frac{a}{11} = \frac{b}{22} = \frac{c}{42}.$$

No other option satisfies this equality; for example, under Option A we would have

$$\frac{c}{21}=16796 \neq 8398=\frac{a}{11}.$$

Hence, the correct answer is Option C.

To solve for the coefficients $$\alpha$$ and $$\beta$$, we can use the binomial identity for the sum of two powers:

$$(a+b)^n + (a-b)^n = 2 \left[ \binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \binom{n}{4}a^{n-4}b^4 + \dots \right]$$

1. Simplification of the Expression

Given the expression $$\left(x + \sqrt{x^2 - 1}\right)^6 + \left(x - \sqrt{x^2 - 1}\right)^6$$, let a = x and $$b=\sqrt{\ x^2-1}$$. Applying the identity with $$n=6$$:

$$E = 2 \left[ \binom{6}{0}x^6 + \binom{6}{2}x^4(x^2-1)^1 + \binom{6}{4}x^2(x^2-1)^2 + \binom{6}{6}(x^2-1)^3 \right]$$

$$E = 2 \left[ x^6 + 15x^4(x^2-1) + 15x^2(x^2-1)^2 + (x^2-1)^3 \right]$$

2. Expanding the Terms

Now, expand the powers of $$\left(x^2-1\right)$$:

• $$15x^4\left(x^2-1\right)\ =\ 15x^6\ -\ 15x^4$$
  
• $$15x^4\left(x^2-1\right)^2\ =\ 15x^6\ -\ 30x^4+15x^2$$
  
• $$\left(x^2-1\right)^3\ =\ x^6-3x^4+3x^2-1$$
  

Substitute these back into the expression:

$$E = 2 \left[ (x^6) + (15x^6 - 15x^4) + (15x^6 - 30x^4 + 15x^2) + (x^6 - 3x^4 + 3x^2 - 1) \right]$$

3. Combining Like Terms

Group the terms by powers of x:

• $$x^6$$ terms: $$1+15+15+1\ =\ 32$$
• $$x^4$$ terms: $$-15-30-3\ =\ -48$$
• $$x^2$$ terms: $$15+3\ =\ 18$$
• Constant: -1

$$E\ =\ 2\left[32x^6\ -\ 48x^4+18x^2-1\right]$$

4. Finding $$\alpha$$ and $$\beta$$

From the expanded form, we identify the coefficients:

• $$\alpha$$  =  -96
• $$\beta$$  =  36

Now, calculate $$\alpha - \beta$$:

$$\alpha - \beta = -96 - 36 = -132$$

Final Answer:

The correct option is D, $$\alpha - \beta = -132$$.

First recall the Binomial Theorem. For any non-negative integer $$N$$, the expansion of $$(1+x)^N$$ is

$$ (1+x)^N \;=\; \sum_{r=0}^{N} {N \choose r}\,x^{r}, $$

where the coefficient of the term in $$x^{r}$$ is the binomial coefficient $$\displaystyle {N \choose r} = \frac{N!}{r!\,(N-r)!}.$$

In the given question the power of the binomial is $$N = n+5,$$ so the general coefficient is $$\displaystyle {\,n+5 \choose r}.$$

We are told that for some positive integer $$n$$ three consecutive coefficients are in the ratio $$5:10:14.$$ Let the three consecutive terms correspond to the powers $$r-1,\; r,\; r+1.$$ Therefore we can set

$$ {N \choose r-1} : {N \choose r} : {N \choose r+1}\;=\;5:10:14. $$

Introduce a common factor $$k$$ (so that all three real coefficients are multiples of it) and write

$$ {N \choose r-1}=5k,\qquad {N \choose r}=10k,\qquad {N \choose r+1}=14k. $$

To use the given ratios conveniently we employ the standard relationship between successive binomial coefficients:

$$ \frac{{N \choose r}}{{N \choose r-1}}=\frac{N-r+1}{r},\qquad\text{and}\qquad \frac{{N \choose r}}{{N \choose r+1}}=\frac{r+1}{N-r}. $$

Now compare the first two coefficients. From the given ratio we have

$$ \frac{{N \choose r-1}}{{N \choose r}}=\frac{5}{10}=\frac12. $$

But the theoretical ratio is

$$ \frac{{N \choose r-1}}{{N \choose r}}=\frac{r}{N-r+1}. $$

Equating these two expressions gives

$$ \frac{r}{N-r+1}=\frac12 \;\Longrightarrow\;2r=N-r+1 \;\Longrightarrow\;3r=N+1 \;\Longrightarrow\;N=3r-1. $$

Next compare the second and the third coefficients. From the given ratio we have

$$ \frac{{N \choose r}}{{N \choose r+1}}=\frac{10}{14}=\frac{5}{7}. $$

The theoretical ratio is

$$ \frac{{N \choose r}}{{N \choose r+1}}=\frac{r+1}{N-r}. $$

Equating these expressions yields

$$ \frac{r+1}{N-r}=\frac{5}{7} \;\Longrightarrow\;7(r+1)=5(N-r) \;\Longrightarrow\;7r+7=5N-5r \;\Longrightarrow\;12r+7=5N. $$

Substitute $$N=3r-1$$ from the first relation into the second:

$$ 12r+7=5(3r-1) \;\Longrightarrow\;12r+7=15r-5 \;\Longrightarrow\;3r=12 \;\Longrightarrow\;r=4. $$

With $$r=4,$$ return to $$N=3r-1$$ to find

$$ N=3(4)-1=12-1=11. $$

Remembering that $$N=n+5,$$ we obtain

$$ n+5=11 \;\Longrightarrow\;n=6. $$

We now know the actual expansion is $$(1+x)^{11}.$$ For $$N=11$$ the binomial coefficients are $$\displaystyle {11 \choose 0}, {11 \choose 1},\ldots,{11 \choose 11}.$$ The coefficients increase up to the middle and then decrease symmetrically. Because 11 is odd, the two central (and equal largest) coefficients occur at

$$ r=\frac{11-1}{2}=5 \quad\text{and}\quad r+1=6. $$

Compute one of them:

$$ {11 \choose 5} =\frac{11!}{5!\,6!} =\frac{11\times10\times9\times8\times7}{5\times4\times3\times2\times1} =462. $$

The coefficient $$\displaystyle {11 \choose 6}$$ equals the same value by symmetry, so the largest coefficient in the entire expansion is $$462.$$

Hence, the correct answer is Option A.

First recall the binomial theorem: for any two terms $$a$$ and $$b$$ and a positive integer $$n$$, the expansion is $$ (a+b)^n=\displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{\,n-r}b^{\,r}\,. $$

Here we have $$a=\sqrt{x}=x^{1/2},\; b=-\dfrac{k}{x^{2}}= -k\,x^{-2},\; n=10.$$

So the general term in the expansion is obtained by choosing the $$r^{\text{th}}$$ term (starting from $$r=0$$):

$$ T_r=\binom{10}{r}\left(\sqrt{x}\right)^{\,10-r}\!\left(-\dfrac{k}{x^{2}}\right)^{r}. $$

We simplify the powers of $$x$$ inside $$T_r$$. Using the laws of indices $$x^{m}\,x^{n}=x^{m+n}$$ and $$(x^{m})^{n}=x^{mn},$$ we write

$$\left(\sqrt{x}\right)^{\,10-r}=x^{\frac{1}{2}(10-r)} \quad\text{and}\quad \left(-\dfrac{k}{x^{2}}\right)^{r}=(-k)^{r}\,x^{-2r}.$$ Therefore

$$ T_r=\binom{10}{r}(-k)^{r}\,x^{\frac{1}{2}(10-r)}\,x^{-2r} =\binom{10}{r}(-k)^{r}\,x^{\,\frac{10-r}{2}-2r}. $$

The exponent of $$x$$ in $$T_r$$ is

$$ \frac{10-r}{2}-2r =5-\frac{r}{2}-2r =5-\frac{5r}{2}. $$

For the term to be constant (i.e., independent of $$x$$) this exponent must be zero:

$$ 5-\frac{5r}{2}=0 \;\Longrightarrow\; \frac{5r}{2}=5 \;\Longrightarrow\; r=2. $$

Thus the constant term comes from $$r=2$$. We now substitute $$r=2$$ back into the expression for $$T_r$$ (noting that $$(-k)^2=k^2$$ because the square removes the minus sign):

$$ T_2=\binom{10}{2}\,k^{2}\,x^{\,\frac{10-2}{2}-2\!\cdot\!2} =\binom{10}{2}\,k^{2}\,x^{\,4-4} =\binom{10}{2}\,k^{2}\,x^{0} =\binom{10}{2}\,k^{2}. $$

Since $$\displaystyle\binom{10}{2}=\dfrac{10\cdot9}{2}=45,$$ we have

$$ T_2 = 45\,k^{2}. $$

The problem states that this constant term equals 405, so

$$ 45\,k^{2}=405 \;\Longrightarrow\; k^{2}= \frac{405}{45}=9. $$

Taking square roots, $$|k|=\sqrt{9}=3.$$

Hence, the correct answer is Option C.

The expression to be expanded is $$\left(\dfrac{3}{2}x^{2}-\dfrac{1}{3x}\right)^{9}$$. We shall use the Binomial Theorem, which states that for any real numbers $$a$$, $$b$$ and a non-negative integer $$n$$, we have $$\displaystyle (a+b)^{n}=\sum_{r=0}^{n}{^{n}C_{r}}\,a^{\,n-r}\,b^{\,r}$$, where $${^{n}C_{r}}=\dfrac{n!}{r!(n-r)!}$$.

Here we identify $$a=\dfrac{3}{2}x^{2}$$ and $$b=-\dfrac{1}{3x}$$, with $$n=9$$. The general term obtained from the expansion is therefore

$$T_{r}={^{9}C_{r}}\left(\dfrac{3}{2}x^{2}\right)^{9-r}\left(-\dfrac{1}{3x}\right)^{r},\qquad r=0,1,2,\dots,9.$$

First we examine the power of $$x$$ in this term. We have

$$\left(\dfrac{3}{2}x^{2}\right)^{9-r} \; \text{contributes } x^{\,2(9-r)},$$ $$\left(-\dfrac{1}{3x}\right)^{r} \; \text{contributes } x^{-\,r}.$$

Multiplying these powers, the total exponent of $$x$$ in $$T_{r}$$ becomes

$$2(9-r) - r \;=\;18 - 3r.$$

We need the term that is independent of $$x$$, so we set this exponent equal to zero:

$$18 - 3r = 0 \quad\Longrightarrow\quad r = 6.$$

Thus the required term is obtained when $$r=6$$. Substituting $$r=6$$ into the general term, we get

$$T_{6} = {^{9}C_{6}}\left(\dfrac{3}{2}x^{2}\right)^{9-6}\left(-\dfrac{1}{3x}\right)^{6}.$$

Simplifying each factor one by one:

$${^{9}C_{6}} = {^{9}C_{3}} = \dfrac{9\times8\times7}{3\times2\times1} = 84,$$ $$\left(\dfrac{3}{2}x^{2}\right)^{3} = \left(\dfrac{3}{2}\right)^{3}x^{6} = \dfrac{27}{8}x^{6},$$ $$\left(-\dfrac{1}{3x}\right)^{6} = (-1)^{6}\dfrac{1}{3^{6}}x^{-6} = \dfrac{1}{729}x^{-6}.$$

Multiplying these three pieces, the powers of $$x$$ cancel out ($$x^{6}\cdot x^{-6}=x^{0}=1$$), and the numerical coefficient becomes

$$k \;=\; 84 \times \dfrac{27}{8} \times \dfrac{1}{729}.$$

Notice that $$729 = 27 \times 27$$, so

$$\dfrac{27}{729} = \dfrac{1}{27}.$$

Therefore

$$k = 84 \times \dfrac{1}{8} \times \dfrac{1}{27} = 84 \times \dfrac{1}{216} = \dfrac{84}{216}.$$

Dividing numerator and denominator by $$12$$ gives

$$k = \dfrac{7}{18}.$$

Finally, we are asked for $$18k$$:

$$18k = 18 \times \dfrac{7}{18} = 7.$$

Hence, the correct answer is Option D.

We recall that for any real number $$x$$, its fractional part is defined by the formula $$\{x\}=x-\lfloor x\rfloor$$, where $$\lfloor x\rfloor$$ is the greatest integer less than or equal to $$x$$.

We need $$\left\{\dfrac{3^{200}}{8}\right\}$$. To isolate the fractional part, it is convenient to know how $$3^{200}$$ behaves when divided by $$8$$. For this purpose we look at remainders, that is, we work “modulo 8”.

We start examining small powers of $$3$$ modulo $$8$$:

$$3^{1}=3\equiv3\pmod{8},$$
$$3^{2}=9\equiv9-8=1\pmod{8}.$$

Now $$3^{3}=3^{2}\cdot3\equiv1\cdot3=3\pmod{8},$$
and $$3^{4}=(3^{2})^{2}\equiv1^{2}=1\pmod{8}.$$

We see the alternating pattern $$3,1,3,1,\dots$$. In particular, every even power of $$3$$ leaves a remainder of $$1$$ upon division by $$8$$. Because $$200$$ is even, we have

$$3^{200}\equiv1\pmod{8}.$$

This congruence means there exists some integer $$k$$ such that

$$3^{200}=8k+1.$$

We now divide both sides by $$8$$:

$$\dfrac{3^{200}}{8}=k+\dfrac{1}{8}.$$

Here $$k$$ is an integer, so by the definition of fractional part,

$$\left\{\dfrac{3^{200}}{8}\right\}=\dfrac{1}{8}.$$

Hence, the correct answer is Option D.

We start with the binomial expansion

$$\left(3^{\tfrac12}+5^{\tfrac18}\right)^n=\sum_{k=0}^{n}\binom{n}{k}\left(3^{\tfrac12}\right)^{\,n-k}\left(5^{\tfrac18}\right)^{\,k}.$$

The general (k-th) term of this expansion is therefore

$$T_k=\binom{n}{k}\,3^{\tfrac{n-k}{2}}\;5^{\tfrac{k}{8}}.$$

For this term to be an integral (i.e., non-fractional) power of both 3 and 5, the exponents of 3 and 5 must be non-negative integers. So we impose two separate conditions:

1. The exponent of 3 is $$(n-k)/2,$$ which must be an integer.

2. The exponent of 5 is $$k/8,$$ which must also be an integer.

Now translate these statements into congruences:

• From $$\dfrac{k}{8}\in\mathbb Z$$ we get $$k\equiv 0\pmod{8}.$$ Hence we may write $$k=8m,$$ where $$m$$ is a non-negative integer.

• From $$\dfrac{n-k}{2}\in\mathbb Z$$ we get $$n-k\equiv 0\pmod{2}.$$ Substituting $$k=8m$$ gives $$n-8m\equiv 0\pmod{2}.$$ Because $$8m$$ is always even, this reduces to $$n\equiv 0\pmod{2},$$ so $$n$$ itself must be even.

Therefore every integer term corresponds to a non-negative integer $$m$$ satisfying $$k=8m\le n.$$ Equivalently, $$m$$ can take all integer values from $$0$$ up to

$$\left\lfloor\frac{n}{8}\right\rfloor.$$

Counting these values, we get exactly

$$\left\lfloor\frac{n}{8}\right\rfloor+1$$

integral terms in the whole expansion.

The statement of the problem tells us that the number of integral terms is $$33.$$ Hence

$$\left\lfloor\frac{n}{8}\right\rfloor+1=33.$$

Subtracting 1 from both sides gives

$$\left\lfloor\frac{n}{8}\right\rfloor=32.$$

This means that $$n/8$$ lies between $$32$$ and $$33$$, but is not equal to $$33$$. In inequality form,

$$32\le\frac{n}{8}\lt 33.$$

Multiplying through by $$8$$, we obtain

$$256\le n\lt264.$$

Because $$n$$ must be an integer, the possible values are $$256,257,258,259,260,261,262,263.$$ But we also discovered earlier that $$n$$ must be even, so the admissible set narrows down to $$256,258,260,262.$$

We are asked for the least value of $$n,$$ which is clearly $$n=256.$$

Hence, the correct answer is Option C.

We have to study the binomial expansion of $$\left(\alpha\,x^{\tfrac19}+\beta\,x^{-\tfrac16}\right)^{10}$$ where $$\alpha\gt 0,\;\beta\gt 0$$ and they satisfy the relation $$\alpha^{3}+\beta^{2}=4\;.$$

The general term (the $$(r+1)$$-th term) of $$(u+v)^{10}$$ is given by the binomial theorem

$$T_{r}=\,{{}^{10}C_{r}}\;u^{\,10-r}\,v^{\,r}\;.$$

Here $$u=\alpha\,x^{\tfrac19},\;v=\beta\,x^{-\tfrac16}\;,$$ so

$$T_{r}={{}^{10}C_{r}}\left(\alpha\,x^{\tfrac19}\right)^{10-r}\!\left(\beta\,x^{-\tfrac16}\right)^{r} ={{}^{10}C_{r}}\;\alpha^{\,10-r}\beta^{\,r}\;x^{\;\tfrac{10-r}{9}-\tfrac{r}{6}}\;.$$

We want the term independent of $$x$$, that is, the exponent of $$x$$ must be zero:

$$\frac{10-r}{9}-\frac{r}{6}=0\;.$$

Multiplying by $$18$$ to clear denominators, we obtain

$$2(10-r)-3r=0\;\Longrightarrow\;20-2r-3r=0\;\Longrightarrow\;20-5r=0\;\Longrightarrow\;r=4\;.$$

Substituting $$r=4$$ in the general term gives the required constant term:

$$T_{4}={{}^{10}C_{4}}\;\alpha^{\,10-4}\beta^{\,4} ={{}^{10}C_{4}}\;\alpha^{6}\beta^{4}\;.$$

We know $${}^{10}C_{4}=210\;,$$ hence

$$T_{4}=210\,\alpha^{6}\beta^{4}\;.$$

To maximise this expression subject to the constraint $$\alpha^{3}+\beta^{2}=4$$, we set

$$u=\alpha^{3}\quad(u\gt 0),\quad v=\beta^{2}\quad(v\gt 0)\;.$$

The given condition becomes

$$u+v=4\;.$$

Also,

$$\alpha^{6}=u^{2},\qquad\beta^{4}=v^{2}\;,$$

so

$$T_{4}=210\,u^{2}v^{2}=210\,(uv)^{2}\;.$$

Under the fixed sum $$u+v=4$$, the product $$uv$$ attains its maximum when $$u=v$$ (AM-GM inequality or elementary calculus). Thus

$$u=v=2\quad\Longrightarrow\quad uv=2\times2=4\;.$$

Therefore the maximum value of the required term is

$$T_{4,\max}=210\,(4)^{2}=210\times16=3360\;.$$

The question states that this maximum equals $$10k$$, so

$$10k=3360\;\Longrightarrow\;k=\frac{3360}{10}=336\;.$$

Hence, the correct answer is Option A.

We are asked to find the coefficient of $$x^7$$ in the long sum

$$$S \;=\; (1 + x)^{10} \;+\; x(1 + x)^{9} \;+\; x^{2}(1 + x)^{8} \;+\; \cdots \;+\; x^{10}.$$$

For convenience we rewrite this expression in sigma notation, so that every term follows the same pattern. Observing the powers, we can write

$$$S \;=\; \sum_{k = 0}^{10} x^{\,k}\,(1 + x)^{\,10 - k}.$$$

Here the index $$k$$ tells us the shift contributed by the factor $$x^k,$$ while $$(1 + x)^{10-k}$$ is the binomial factor that is still to be expanded.

Our goal is to collect the total coefficient of $$x^7$$ in this entire sum. A term from the $$k^{\text{th}}$$ summand will have the general form

$$$x^{\,k}\,\bigl[\text{coefficient of }x^{\,j}\text{ in }(1 + x)^{10 - k}\bigr] \;=\; x^{\,k}\,\binom{\,10 - k\,}{\,j\,}x^{\,j} \;=\; \binom{\,10 - k\,}{\,j\,}x^{\,k + j}.$$$

To make the power of $$x$$ equal to 7, the indices must satisfy

$$k + j \;=\; 7.$$

Because $$j$$ must be a non-negative integer not exceeding $$10 - k,$$ only those $$k$$ from 0 up to 7 are possible; for each such $$k$$ we then have the unique choice $$j = 7 - k.$$ Substituting $$j = 7 - k$$ in the binomial coefficient gives the contribution from the $$k^{\text{th}}$$ term:

$$$\text{Contribution}(k) \;=\; \binom{\,10 - k\,}{\,7 - k\,}.$$$

We now list every admissible $$k$$ and compute the corresponding binomial coefficient one by one, showing every arithmetic step.

For $$k = 0$$: $$\binom{10}{7} = \frac{10!}{7!\,3!} = 120.$$

For $$k = 1$$: $$\binom{9}{6} = \frac{9!}{6!\,3!} = 84.$$

For $$k = 2$$: $$\binom{8}{5} = \frac{8!}{5!\,3!} = 56.$$

For $$k = 3$$: $$\binom{7}{4} = \frac{7!}{4!\,3!} = 35.$$

For $$k = 4$$: $$\binom{6}{3} = \frac{6!}{3!\,3!} = 20.$$

For $$k = 5$$: $$\binom{5}{2} = \frac{5!}{2!\,3!} = 10.$$

For $$k = 6$$: $$\binom{4}{1} = 4.$$

For $$k = 7$$: $$\binom{3}{0} = 1.$$

All remaining $$k$$ (namely 8, 9, 10) are not allowed because they would require $$j$$ to be negative, which is impossible in a binomial expansion. Now we add up all the contributions obtained:

$$$\begin{aligned} \text{Total coefficient} &= 120 + 84 + 56 + 35 + 20 + 10 + 4 + 1 \\ &= 204 + 56 + 35 + 20 + 10 + 4 + 1 \\ &= 260 + 35 + 20 + 10 + 4 + 1 \\ &= 295 + 20 + 10 + 4 + 1 \\ &= 315 + 10 + 4 + 1 \\ &= 325 + 4 + 1 \\ &= 329 + 1 \\ &= 330. \end{aligned}$$$

So, the coefficient of $$x^7$$ in the given sum is $$330$$.

Hence, the correct answer is Option B.

We have to find the numerical value of the finite sum

$$\sum_{r=0}^{20} {^{\,50-r}\!C_{\,6}}.$$

The upper index of the combination depends on $$r$$, so first we replace that dependent index by a single symbol that increases in the usual (left-to-right) direction. Put

$$k = 50 - r.$$

When $$r = 0$$, we obtain $$k = 50$$, and when $$r = 20$$, we obtain $$k = 30$$. As $$r$$ runs from $$0$$ to $$20$$, $$k$$ therefore runs from $$50$$ down to $$30$$. Because addition is commutative, the order does not matter, so we may rewrite the sum as

$$\sum_{r=0}^{20} {^{\,50-r}\!C_{\,6}} \;=\; \sum_{k=30}^{50} {^{\,k}\!C_{\,6}}.$$

Now we recall a standard combinatorial identity, usually proved by induction or by Pascal’s rule:

$$\sum_{k=m}^{n} {^{\,k}\!C_{\,m}} \;=\; {^{\,n+1}\!C_{\,m+1}} \quad\text{for integers } n \ge m \ge 0.$$

In our case $$m = 6$$, so if the lower limit were $$6$$ instead of $$30$$, we could apply the identity directly. We therefore write the desired sum as the difference of two larger sums whose lower limit is $$6$$:

$$\sum_{k=30}^{50} {^{\,k}\!C_{\,6}} \;=\; \left(\sum_{k=6}^{50} {^{\,k}\!C_{\,6}}\right) \;-\; \left(\sum_{k=6}^{29} {^{\,k}\!C_{\,6}}\right).$$

By the quoted identity, we evaluate each parenthesis separately.

First parenthesis (upper limit $$50$$):

$$\sum_{k=6}^{50} {^{\,k}\!C_{\,6}} = {^{\,50+1}\!C_{\,6+1}} = {^{\,51}\!C_{\,7}}.$$

Second parenthesis (upper limit $$29$$):

$$\sum_{k=6}^{29} {^{\,k}\!C_{\,6}} = {^{\,29+1}\!C_{\,6+1}} = {^{\,30}\!C_{\,7}}.$$

Substituting these two evaluations back, we get

$$\sum_{k=30}^{50} {^{\,k}\!C_{\,6}} = {^{\,51}\!C_{\,7}} \;-\; {^{\,30}\!C_{\,7}}.$$

Hence, the original sum equals $${}^{51}C_7 - {}^{30}C_7$$, which coincides with Option A.

Hence, the correct answer is Option A.

We have the binomial

$$\left(\dfrac{x}{\cos\theta}+\dfrac{1}{x\sin\theta}\right)^{16}$$

and we want the term that is independent of $$x$$. For a binomial $$(a+b)^n$$ the general term is, by the binomial theorem,

$$T_r=\;{^{n}C_{r}}\;a^{\,n-r}\,b^{\,r},\qquad r=0,1,2,\ldots ,n.$$

Here $$n=16,\;a=\dfrac{x}{\cos\theta},\;b=\dfrac{1}{x\sin\theta}.$$ So the general term is

$$T_r={^{16}C_{r}}\left(\dfrac{x}{\cos\theta}\right)^{16-r} \left(\dfrac{1}{x\sin\theta}\right)^{r}.$$

Combining the powers of $$x$$:

$$T_r={^{16}C_{r}}\;x^{(16-r)}\,x^{-r}\;\cos^{-\,\!(16-r)}\!\theta\,\sin^{-\,\!r}\!\theta ={^{16}C_{r}}\;x^{\,16-2r}\;(\cos\theta)^{-(16-r)}(\sin\theta)^{-r}.$$

The term is independent of $$x$$ when the exponent of $$x$$ is zero, i.e.

$$16-2r=0\;\Longrightarrow\;r=8.$$

Hence there is a single $$x$$-free term, obtained by putting $$r=8$$:

$$T_{\text{indep}}={^{16}C_{8}} \;(\cos\theta)^{-(16-8)}(\sin\theta)^{-8} ={^{16}C_{8}}\;(\cos\theta)^{-8}(\sin\theta)^{-8} ={^{16}C_{8}}\;(\sin\theta\cos\theta)^{-8}.$$

Thus, for every allowed $$\theta,$$

$$\boxed{\,l(\theta)={^{16}C_{8}}\,[\sin\theta\cos\theta]^{-8}\,}.$$

To find the least value, we need the maximum of $$\sin\theta\cos\theta$$ in the given interval because the expression is the reciprocal of its eighth power. Using the identity

$$\sin\theta\cos\theta=\dfrac{1}{2}\sin2\theta,$$

we note that in the first quadrant $$\sin2\theta$$ (and hence $$\sin\theta\cos\theta$$) increases with $$\theta$$ until $$\theta=\dfrac{\pi}{4}.$$ Therefore:

First interval $$\dfrac{\pi}{8}\le\theta\le\dfrac{\pi}{4}:$$ The maximum occurs at the right end, $$\theta=\dfrac{\pi}{4}.$$ Then

$$\sin\dfrac{\pi}{4}=\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2},\qquad \sin\dfrac{\pi}{4}\cos\dfrac{\pi}{4}=\dfrac12.$$

Hence

$$l_1={^{16}C_{8}}\left(\dfrac12\right)^{-8} ={^{16}C_{8}}\;2^{8} ={^{16}C_{8}}\times256.$$

Second interval $$\dfrac{\pi}{16}\le\theta\le\dfrac{\pi}{8}:$$ Again the maximum is at the right end, $$\theta=\dfrac{\pi}{8}.$$ Now

$$\sin\frac{\pi}{8}\cos\frac{\pi}{8} =\frac12\sin\frac{\pi}{4} =\frac12\cdot\frac{\sqrt2}{2} =\frac{\sqrt2}{4}.$$

Observe that $$\dfrac{\sqrt2}{4}=2^{-3/2},$$ so

$$(\sin\theta\cos\theta)^{8}=2^{-12}\quad\text{and}\quad (\sin\theta\cos\theta)^{-8}=2^{12}=4096.$$

Therefore

$$l_2={^{16}C_{8}}\times4096.$$

Finally, the required ratio is

$$\frac{l_2}{l_1} =\frac{{^{16}C_{8}}\times4096}{{^{16}C_{8}}\times256} =\frac{4096}{256} =16.$$

So $$l_2:l_1=16:1.$$

Hence, the correct answer is Option 2.

We start with the binomial expression

$$\left(1+\dfrac1x\right)^n.$$

The binomial theorem states that

$$\left(1+\dfrac1x\right)^n=\sum_{k=0}^{n}\binom{n}{k}\,1^{\,n-k}\left(\dfrac1x\right)^{k}=\sum_{k=0}^{n}\binom{n}{k}\,x^{-k}.$$

The exponent of $$x$$ in each term is $$-k$$, which goes from $$-n$$ up to $$0$$. Hence, when the terms are written in increasing powers of $$x$$ (that is, from the most negative exponent to the least negative exponent), the coefficients appear in the order

$$\binom{n}{n},\;\binom{n}{n-1},\;\binom{n}{n-2},\;\dots,\;\binom{n}{1},\;\binom{n}{0}.$$

Thus three consecutive coefficients in this order are

$$\binom{n}{r},\;\binom{n}{r-1},\;\binom{n}{r-2}$$

for some integer $$r$$ with $$2\le r\le n.$$ These three coefficients are given to be in the ratio $$2:5:12$$. Therefore we must have

$$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{2}{5}\qquad\text{and}\qquad \frac{\binom{n}{r-1}}{\binom{n}{r-2}}=\frac{5}{12}.$$

To evaluate these ratios, we use the standard relation for consecutive binomial coefficients:

$$\frac{\binom{n}{k}}{\binom{n}{k-1}}=\frac{n-k+1}{k}.$$ Setting $$k=r$$ gives

$$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{n-r+1}{r}.$$

Setting $$k=r-1$$ gives

$$\frac{\binom{n}{r-1}}{\binom{n}{r-2}}=\frac{n-(r-1)+1}{\,r-1\,}=\frac{n-r+2}{r-1}.$$

Now we equate these expressions to the given ratios.

First ratio:

$$\frac{n-r+1}{r}=\frac{2}{5}.$$

Cross-multiplying, we obtain

$$5(n-r+1)=2r,$$ $$5n-5r+5=2r,$$ $$5n+5=7r,$$ $$r=\frac{5n+5}{7}. \quad -(1)$$

Second ratio:

$$\frac{n-r+2}{r-1}=\frac{5}{12}.$$

Cross-multiplying, we obtain

$$12(n-r+2)=5(r-1),$$ $$12n-12r+24=5r-5,$$ $$12n+24=17r-5,$$ $$12n+29=17r,$$ $$r=\frac{12n+29}{17}. \quad -(2)$$

Since both (1) and (2) equal $$r$$, we equate them:

$$\frac{5n+5}{7}=\frac{12n+29}{17}.$$

Cross-multiplying gives

$$17(5n+5)=7(12n+29),$$ $$85n+85=84n+203.$$

Moving like terms to the same side, we get

$$85n-84n=203-85,$$ $$n=118.$$

Because $$n=118$$ is a positive integer and satisfies both equations, it is the required value.

Hence, the correct answer is Option 118.

We have to study the product

$$P(x)=(1+x+x^2+\ldots +x^{2n})\,(1-x+x^2-x^3+\ldots +x^{2n}).$$

Denote the two factors separately as

$$A(x)=1+x+x^2+\ldots +x^{2n},\qquad B(x)=1-x+x^2-x^3+\ldots +x^{2n}.$$

The problem asks for the sum of the coefficients of all even powers of $$x$$ in $$P(x)$$, and tells us that this sum equals 61. A classical trick allows us to obtain that sum without expanding the whole product.

For any polynomial $$Q(x)=q_0+q_1x+q_2x^2+\ldots,$$ the sum of the coefficients of its even-powered terms is

$$\text{(sum of even coefficients)}=\frac{Q(1)+Q(-1)}{2}.$$

This identity holds because substituting $$x=1$$ totals all coefficients, while substituting $$x=-1$$ subtracts every odd-powered coefficient from its even neighbour, leaving only twice the even-powered ones. Dividing by 2 therefore recovers the desired sum.

Applying that idea to our polynomial $$P(x)=A(x)B(x),$$ we need $$P(1)$$ and $$P(-1).$$ We compute them factor by factor.

First evaluate each factor at $$x=1$$:

$$A(1)=1+1+1+\ldots +1\quad\text{(total of }2n+1\text{ ones)}=2n+1.$$

For $$B(1)$$ we have an alternating sum with an even number of sign changes, because the highest power is $$x^{2n}$$ where $$2n$$ is even. Writing the first few terms makes the pattern clear:

$$B(1)=1-1+1-1+\ldots -1+1.$$

Every adjacent pair $$1-1$$ cancels to 0, and since the total number of terms is odd (namely $$2n+1$$), one lone $$+1$$ remains at the end. Therefore

$$B(1)=1.$$

Hence

$$P(1)=A(1)\,B(1)=(2n+1)\times 1=2n+1.$$

Next evaluate at $$x=-1$$. Begin with $$A(-1)$$:

$$A(-1)=1+(-1)+(-1)^2+(-1)^3+\ldots +(-1)^{2n}.$$

This is the same alternating sum just seen, and, as before, the terms cancel pairwise leaving a single $$+1$$, so

$$A(-1)=1.$$

For $$B(-1)$$ every term picks up a factor $$(-1)^k$$ from the variable and already carries the coefficient $$(-1)^k$$, giving

$$B(-1)=\sum_{k=0}^{2n}(-1)^k\,(-1)^k=\sum_{k=0}^{2n}(-1)^{2k}=\sum_{k=0}^{2n}1.$$

Because $$(-1)^{2k}=1$$ for every integer $$k$$, we simply get $$2n+1$$ ones:

$$B(-1)=2n+1.$$

Therefore

$$P(-1)=A(-1)\,B(-1)=1\times(2n+1)=2n+1.$$

With both values ready, we invoke the even-coefficient formula stated earlier:

$$\text{Sum of even coefficients}=\frac{P(1)+P(-1)}{2} =\frac{(2n+1)+(2n+1)}{2}=2n+1.$$

The question tells us that this sum equals 61, so we write

$$2n+1=61.$$

Solving for $$n$$ gives

$$2n=60\quad\Longrightarrow\quad n=30.$$

Hence, the correct answer is Option 30.

We have to compare the coefficients of $$x^7$$ and $$x^{13}$$ in the expansion of $$(2x^2+3x+4)^{10}$$.

First recall the multinomial theorem, which states that for any positive integer $$n$$,

$$ (a+b+c)^n=\sum_{k_1+k_2+k_3=n}\dfrac{n!}{k_1!k_2!k_3!}\;a^{\,k_1}b^{\,k_2}c^{\,k_3}. $$

Here we identify

$$ a=2x^2,\qquad b=3x,\qquad c=4,\qquad n=10. $$

So the general term after expansion is

$$ \dfrac{10!}{k_1!k_2!k_3!}\,(2x^2)^{k_1}\,(3x)^{k_2}\,4^{\,k_3}, $$

where the exponents satisfy $$k_1+k_2+k_3=10$$ with each $$k_i\ge 0$$. This general term contributes

$$ \dfrac{10!}{k_1!k_2!k_3!}\,2^{k_1}3^{k_2}4^{k_3}\;x^{\,2k_1+k_2}. $$

Hence the power of $$x$$ in any term is determined by

$$ 2k_1+k_2=r. $$

We shall solve this relation twice, once for $$r=7$$ and once for $$r=13$$.

Case 1: $$r=7$$. Setting $$2k_1+k_2=7$$ with the additional condition $$k_1+k_2+k_3=10$$, we write $$k_1=t$$. Then

$$ k_2=7-2t,\qquad k_3=10-t-(7-2t)=3+t. $$

Non-negative integers give the admissible values

$$ t=0,1,2,3. $$

Consequently the four ordered triples are

$$ (k_1,k_2,k_3)=(0,7,3),(1,5,4),(2,3,5),(3,1,6). $$

Thus

$$ a_7=\sum_{t=0}^{3}\dfrac{10!}{k_1!k_2!k_3!}\;2^{k_1}3^{k_2}4^{k_3}. $$

Case 2: $$r=13$$. Now $$2k_1+k_2=13$$. Put again $$k_1=t$$; then

$$ k_2=13-2t,\qquad k_3=10-t-(13-2t)=t-3. $$

We require $$k_3\ge 0,$$ so $$t\ge 3$$. Allowable integer values are

$$ t=3,4,5,6. $$

This yields another set of four triples:

$$ (k_1,k_2,k_3)=(3,7,0),(4,5,1),(5,3,2),(6,1,3). $$

Therefore

$$ a_{13}=\sum_{t=3}^{6}\dfrac{10!}{k_1!k_2!k_3!}\;2^{k_1}3^{k_2}4^{k_3}. $$

Pairwise comparison of the two sets. Notice that each triple in the $$a_{13}$$ list is obtained from the corresponding triple in the $$a_7$$ list by the simple shift

$$ (k_1,k_2,k_3)\longrightarrow(k_1+3,\;k_2,\;k_3-3). $$

Let us compare one matched pair. Take a triple $$(k_1,k_2,k_3)$$ from the $$a_7$$ list and build the twin $$\bigl(k_1+3,k_2,k_3-3\bigr)$$ for $$a_{13}$$. The ratio of their individual contributions is

$$$ \frac{\displaystyle \dfrac{10!}{k_1!k_2!k_3!}\;2^{k_1}3^{k_2}4^{k_3}} {\displaystyle \dfrac{10!}{(k_1+3)!\,k_2!\,(k_3-3)!}\;2^{k_1+3}3^{k_2}4^{\,k_3-3}} = \frac{(k_1+3)!\,(k_3-3)!}{k_1!\,k_3!}\;\frac{2^{k_1}}{2^{k_1+3}}\;\frac{4^{k_3}}{4^{k_3-3}}. $$$

Now we simplify each factor:

• The factorial quotient is

$$\frac{(k_1+3)!\,(k_3-3)!}{k_1!\,k_3!}=\frac{(k_1+3)(k_1+2)(k_1+1)}{k_3(k_3-1)(k_3-2)}.$$

Because for every triple in the $$a_7$$ list we actually have $$k_3=k_1+3,$$ the numerator and the denominator above are identical, giving this whole quotient the value $$1$$.

• The powers of $$2$$ contribute $$2^{-3}=1/8.$$

• The powers of $$4$$ contribute $$4^{3}=64.$$

Multiplying these three simplified factors we obtain

$$ 1\times\frac{1}{8}\times 64=8. $$

This computation shows that each individual term present in $$a_7$$ is exactly eight times the corresponding term in $$a_{13}$$. Since both coefficients are sums of four such matched terms, the same factor $$8$$ carries over to the entire sums. Therefore

$$ \frac{a_7}{a_{13}}=8. $$

So, the answer is $$8$$.

Let us first simplify the expression inside the parentheses by factoring it. We notice that:

$$1 + x + x^2 + x^3 = (1 + x) + x^2(1 + x) = (1 + x)(1 + x^2)$$

Therefore, we can rewrite the entire expression raised to the power of 6 as:

$$(1 + x + x^2 + x^3)^6 = \left[(1 + x)(1 + x^2)\right]^6 = (1 + x)^6(1 + x^2)^6$$

Now, we expand each of these two components using the Binomial Theorem:

$$(1 + x)^6 = \binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \binom{6}{4}x^4 + \dots$$

$$(1 + x^2)^6 = \binom{6}{0} + \binom{6}{1}x^2 + \binom{6}{2}x^4 + \dots$$

Let us evaluate the specific combinations (binomial coefficients) needed for our calculations:

$$\binom{6}{0} = 1$$

$$\binom{6}{1} = 6$$

$$\binom{6}{2} = \frac{6 \times 5}{2} = 15$$

$$\binom{6}{3} = \frac{6 \times 5 \times 4}{6} = 20$$

$$\binom{6}{4} = \binom{6}{2} = 15$$

Substituting these values back into our expansions yields:

$$(1 + x)^6 = 1 + 6x + 15x^2 + 20x^3 + 15x^4 + \dots$$

$$(1 + x^2)^6 = 1 + 6x^2 + 15x^4 + \dots$$

We want to find the coefficient of $$x^4$$ in the product of these two series:

$$\left(1 + 6x + 15x^2 + 20x^3 + 15x^4 + \dots\right) \times \left(1 + 6x^2 + 15x^4 + \dots\right)$$

Let us collect all combinations of terms from the first and second parentheses whose powers of $$x$$ add up to exactly 4:

1) The $$x^4$$ term from the first parentheses multiplied by the constant term ($$1$$) from the second:

$$15x^4 \times 1 = 15x^4$$

2) The $$x^2$$ term from the first parentheses multiplied by the $$x^2$$ term from the second:

$$15x^2 \times 6x^2 = 90x^4$$

3) The constant term ($$1$$) from the first parentheses multiplied by the $$x^4$$ term from the second:

$$1 \times 15x^4 = 15x^4$$

Now, we sum the coefficients of these three cases together to get the total coefficient of $$x^4$$:

$$\text{Coefficient of } x^4 = 15 + 90 + 15 = 120$$

We are asked to evaluate the expression

$$S \;=\; C_0 \;+\; 5\,C_1 \;+\; 9\,C_2 \;+\;\cdots\;+\;(101)\,C_{25},$$

where $$C_r \equiv {}^{25}C_r$$ denotes the binomial coefficient “25 choose r”.

First notice the pattern of the numerical multipliers. Writing out the first few terms we have

$$1 = 4\cdot 0 + 1,\quad 5 = 4\cdot 1 + 1,\quad 9 = 4\cdot 2 + 1,\quad\ldots,\quad 101 = 4\cdot 25 + 1.$$

Hence the general term is $$4r + 1$$ when the binomial coefficient is $$C_r$$. Therefore we can rewrite the sum compactly as

$$S \;=\;\sum_{r=0}^{25} (4r + 1)\,C_r.$$

Now we split the factor $$(4r + 1)$$ into two simpler pieces:

$$4r + 1 \;=\; 1 \;+\; 4r.$$

Substituting this back into the summation gives

$$S \;=\;\sum_{r=0}^{25} 1\cdot C_r \;+\; 4\sum_{r=0}^{25} r\,C_r.$$

We shall evaluate the two sums separately.

1. The first sum $$\displaystyle\sum_{r=0}^{25} C_r$$ is the well-known expansion of $$(1 + 1)^{25}$$ by the binomial theorem. Explicitly, the theorem states

$$(x + y)^n \;=\;\sum_{r=0}^{n} {}^{n}C_r\,x^{n-r}y^{\,r}.$$

Putting $$x = 1,\;y = 1,\;n = 25$$ gives

$$\sum_{r=0}^{25} C_r \;=\;(1 + 1)^{25} \;=\;2^{25}.$$

So

$$\sum_{r=0}^{25} C_r = 2^{25}.$$

2. The second sum $$\displaystyle\sum_{r=0}^{25} r\,C_r$$ can be obtained by differentiating the binomial expansion. Start with $$(1 + x)^{25} = \sum_{r=0}^{25} {}^{25}C_r x^{\,r}.$$ Differentiate both sides with respect to $$x$$:

$$25(1 + x)^{24} = \sum_{r=1}^{25} r\,{}^{25}C_r x^{\,r-1}.$$

Now set $$x = 1$$ to obtain

$$25(1 + 1)^{24} = \sum_{r=1}^{25} r\,C_r\,1^{\,r-1}.$$

Simplifying,

$$25 \cdot 2^{24} = \sum_{r=1}^{25} r\,C_r.$$

(The $$r = 0$$ term is zero anyway, so the upper limit could remain 25.) Thus we have the standard identity

$$\sum_{r=0}^{25} r\,C_r = 25\cdot 2^{24}.$$

Returning to our expression for $$S$$, we now substitute these two evaluated sums:

$$S \;=\; \left(\sum_{r=0}^{25} C_r\right) \;+\; 4\left(\sum_{r=0}^{25} r\,C_r\right) \;=\; 2^{25} \;+\; 4\bigl(25\cdot 2^{24}\bigr).$$

Compute the second term carefully:

$$4\bigl(25\cdot 2^{24}\bigr) \;=\; 100\cdot 2^{24}.$$

Hence

$$S \;=\; 2^{25} \;+\; 100\cdot 2^{24}.$$

Factor out the common power $$2^{24}$$:

$$S \;=\; 2^{24}\bigl(2 + 100\bigr) \;=\; 2^{24}\cdot 102.$$

We are told in the question that $$S = 2^{25}\,k.$$ Therefore equate the two expressions for $$S$$:

$$2^{24}\cdot 102 \;=\; 2^{25}\,k.$$

Divide both sides by $$2^{25}$$ to isolate $$k$$:

$$k \;=\;\frac{2^{24}\cdot 102}{2^{25}} \;=\;\frac{102}{2} \;=\;51.$$

So, the answer is $$51$$.

We have to find the coefficient of $$x^4$$ in the expansion of $$(1 + x + x^2)^{10}$$.

The expression $$(1 + x + x^2)^{10}$$ has three terms being raised to the 10th power. The appropriate tool is the multinomial theorem, which states:

For $$(a + b + c)^{n}$$, a general term is $$\dfrac{n!}{n_1!\,n_2!\,n_3!}\;a^{\,n_1} b^{\,n_2} c^{\,n_3}$$ where $$n_1 + n_2 + n_3 = n$$.

Here we identify $$a = 1,\; b = x,\; c = x^2,\; n = 10$$. After substitution, a general term becomes

$$\dfrac{10!}{n_1!\,n_2!\,n_3!}\;1^{\,n_1} x^{\,n_2} (x^2)^{\,n_3} \;=\; \dfrac{10!}{n_1!\,n_2!\,n_3!}\;x^{\,n_2 + 2n_3}.$$

We need the power of $$x$$ to be exactly 4, so we set

$$n_2 + 2n_3 = 4.$$

Because $$n_1 + n_2 + n_3 = 10$$ and all $$n_i$$ are non-negative integers, we solve $$n_2 + 2n_3 = 4$$ for possible pairs $$(n_2, n_3)$$:

• If $$n_3 = 0$$ then $$n_2 = 4$$, giving $$n_1 = 10 - 4 - 0 = 6.$$br • If $$n_3 = 1$$ then $$n_2 = 2$$, giving $$n_1 = 10 - 2 - 1 = 7.$$br • If $$n_3 = 2$$ then $$n_2 = 0$$, giving $$n_1 = 10 - 0 - 2 = 8.$$

No other non-negative solutions exist, so there are three contributing terms.

Now we calculate the coefficient for each set $$(n_1, n_2, n_3)$$ using the multinomial coefficient $$\dfrac{10!}{n_1!\,n_2!\,n_3!}$$.

For $$(6,4,0):\; \dfrac{10!}{6!\,4!\,0!} =\dfrac{10\times9\times8\times7}{4\times3\times2\times1} =210.$$

For $$(7,2,1):\; \dfrac{10!}{7!\,2!\,1!} =\dfrac{10\times9\times8}{2\times1} =360.$$

For $$(8,0,2):\; \dfrac{10!}{8!\,0!\,2!} =\dfrac{10\times9}{2\times1} =45.$$

Adding all these contributions:

$$210 + 360 + 45 = 615.$$

So, the answer is $$615$$.

We look at the binomial expansion of $$\left(x^m+\dfrac{1}{x^{2}}\right)^{22}$$.

The general ((k+1)-th) term in this expansion is obtained from the formula $$T_{k+1}=^{22}C_{k}\,(x^{m})^{\,22-k}\left(\dfrac{1}{x^{2}}\right)^{k}.$$

Simplifying the powers of $$x$$ we get $$T_{k+1}=^{22}C_{k}\,x^{\,m(22-k)}\;x^{-2k}=^{22}C_{k}\,x^{\,m(22-k)-2k}.$$

We are told that the coefficient of $$x^{1}$$ (i.e. of the term in which the power of $$x$$ is $$1$$) equals $$1540$$. Hence we must have

Power condition: $$m(22-k)-2k=1.$$

Re-arranging, $$m=\dfrac{1+2k}{22-k}$$ where $$k$$ can take integral values from $$0$$ to $$22$$.

For $$m$$ to be a natural number, the denominator $$22-k$$ must divide the numerator $$1+2k$$ exactly. Put $$d=22-k\;(\Rightarrow k=22-d)$$ with $$d=1,2,\dots ,22$$. Substituting,

$$m=\dfrac{1+2(22-d)}{d}=\dfrac{45-2d}{d}=\,\dfrac{45}{d}-2.$$

Thus $$d$$ has to be a positive divisor of $$45$$ not exceeding $$22$$. The admissible values of $$d$$ and the corresponding $$m$$ are

$$ \begin{aligned} d=1&:;; m=45-2=43,\\ d=3&:;; m=15-2=13,\\ d=5&:;; m=9-2=7,\\ d=9&:;; m=5-2=3,\\ d=15&:;; m=3-2=1. \end{aligned} $$

We now check the coefficient of each candidate term, remembering that the coefficient is $$^{22}C_{k}$$ with $$k=22-d$$:

$$ \begin{aligned} d=1&\Rightarrow k=21:\;^{22}C_{21}=^{22}C_{1}=22,\\ d=3&\Rightarrow k=19:\;^{22}C_{19}=^{22}C_{3}=1540,\\ d=5&\Rightarrow k=17:\;^{22}C_{17}=^{22}C_{5}=26334,\\ d=9&\Rightarrow k=13:\;^{22}C_{13}=497420,\\ d=15&\Rightarrow k=7:\;^{22}C_{7}=170544. \end{aligned} $$

The required coefficient is $$1540$$, which occurs only when $$d=3$$ (equivalently $$k=19$$). For this value we have already found $$m=13$$.

So, the answer is $$13$$.

We are required to simplify the expression $$\displaystyle \sum_{r=0}^{25}\;{^{50}C_r}\;{^{50-r}C_{25-r}}$$ and compare it with $$K\;{^{50}C_{25}}$$ in order to determine the constant $$K$$.

First, recall the following standard combinatorial identity, which we will state explicitly before using it:

For any non-negative integers $$n,\;k,\;r$$ with $$0\le r\le k\le n$$, we have $$^{n}C_{r}\;^{\,n-r}C_{k-r}=^{n}C_{k}\;^{k}C_{r}.$$

To see why this holds, write every factorial in the binomial coefficients:

$$^{n}C_{r}\;^{\,n-r}C_{k-r}=\frac{n!}{r!(n-r)!}\;\cdot\;\frac{(n-r)!}{(k-r)!\,(n-k)!} =\frac{n!}{r!\,(k-r)!\,(n-k)!}.$$

On the other hand, $$^{n}C_{k}\;^{k}C_{r}= \frac{n!}{k!(n-k)!}\;\cdot\;\frac{k!}{r!(k-r)!} =\frac{n!}{r!\,(k-r)!\,(n-k)!},$$ which is the same expression. Hence the identity is proved and can be used safely.

Now we apply this identity with the specific values $$n=50$$ and $$k=25$$. For each term of the given sum we therefore have

$$^{50}C_{r}\;^{\,50-r}C_{25-r}=^{50}C_{25}\;^{25}C_{r}.$$

Substituting this result into the entire summation, we obtain

$$\sum_{r=0}^{25}{^{50}C_{r}}\;{^{50-r}C_{25-r}} =\sum_{r=0}^{25}\Bigl(^{50}C_{25}\;^{25}C_{r}\Bigr).$$

The constant $$^{50}C_{25}$$ is independent of the index $$r$$ and can be pulled outside the summation sign:

$$\sum_{r=0}^{25}{^{50}C_{r}}\;{^{50-r}C_{25-r}} =^{50}C_{25}\;\sum_{r=0}^{25}{^{25}C_{r}}.$$

Next, recall another well-known result: the sum of all binomial coefficients of order $$n$$ is a power of two. Explicitly,

$$\sum_{r=0}^{n}{^{\,n}C_{r}} = 2^{\,n}.$$

Here $$n=25$$, so

$$\sum_{r=0}^{25}{^{25}C_{r}} = 2^{25}.$$

Substituting this back into our expression gives

$$\sum_{r=0}^{25}{^{50}C_{r}}\;{^{50-r}C_{25-r}} =^{50}C_{25}\;\bigl(2^{25}\bigr).$$

By direct comparison with the form $$K\;^{50}C_{25}$$, we see that the multiplicative constant must be

$$K = 2^{25}.$$

Hence, the correct answer is Option A.

First we rewrite the given expression in a form that makes every power of $$x$$ visible. We have

$$x^{2}\Bigl(\sqrt{x}+\dfrac{\lambda}{x^{2}}\Bigr)^{10} = x^{2}\Bigl(x^{1/2}+ \lambda\,x^{-2}\Bigr)^{10}.$$

To expand $$(A+B)^{10}$$ we use the Binomial Theorem, stated as $$ (A+B)^{10}= \sum_{r=0}^{10} {10\choose r} A^{\,10-r} B^{\,r}. $$ Here $$A = x^{1/2},\qquad B = \lambda\,x^{-2}.$$

So each general term of the expansion of $$\bigl(x^{1/2}+ \lambda\,x^{-2}\bigr)^{10}$$ is

$$ {10\choose r}\,(x^{1/2})^{\,10-r}\,(\lambda\,x^{-2})^{\,r} = {10\choose r}\,\lambda^{\,r}\,x^{\,(10-r)\cdot\frac12}\,x^{-2r}. $$

Simplifying the exponents of $$x$$ inside that term, we multiply the two powers of $$x$$: $$(10-r)\cdot\frac12 = \frac{10-r}{2},$$ so the total power of $$x$$ in the term (before multiplying by the outer $$x^{2}$$) is

$$\frac{10-r}{2}\;+\;(-2r)=\frac{10-r-4r}{2}=\frac{10-5r}{2}.$$

Now we multiply the whole expansion by the prefactor $$x^{2}$$ that sits outside the bracket. Hence, after multiplication, the power of $$x$$ in the general term becomes

$$ 2+\frac{10-5r}{2}= \frac{4+10-5r}{2}= \frac{14-5r}{2}. $$

We need the coefficient of $$x^{2}$$, so we set this power equal to 2:

$$ \frac{14-5r}{2}=2 \;\Longrightarrow\; 14-5r=4 \;\Longrightarrow\; 5r = 10 \;\Longrightarrow\; r = 2. $$

Thus only the term with $$r=2$$ contributes to the $$x^{2}$$ term.

Substituting $$r=2$$ back into the coefficient part of the term, we get

Coefficient $$= {10\choose 2}\,\lambda^{2} = \frac{10\times9}{2}\,\lambda^{2}=45\,\lambda^{2}.$$

The problem states that this coefficient equals 720, so

$$45\,\lambda^{2}=720 \;\Longrightarrow\; \lambda^{2}= \frac{720}{45}=16 \;\Longrightarrow\; \lambda = 4,$$

and we keep the positive value because the question asks for a positive $$\lambda$$.

Hence, the correct answer is Option C.

We begin with the given summation

$$S \;=\;\sum_{k=1}^{20} k^{2}\,{}^{20}C_{k}.$$

To evaluate this, we recall the binomial expansion formula

$$\bigl(1+x\bigr)^{n} \;=\;\sum_{k=0}^{n} {}^{\,n}C_{k}\,x^{k}.$$

From this, two standard derivative identities follow.

1. Differentiating once with respect to $$x$$ gives

$$n(1+x)^{\,n-1} \;=\;\sum_{k=0}^{n} k\,{}^{n}C_{k}\,x^{\,k-1}.$$

Multiplying both sides by $$x$$ we get the useful form

$$\sum_{k=0}^{n} k\,{}^{n}C_{k}\,x^{\,k} \;=\; n\,x\,(1+x)^{\,n-1}.$$

2. Differentiating the identity in step 1 once more and again multiplying by $$x$$ yields

$$\sum_{k=0}^{n} k(k-1)\,{}^{n}C_{k}\,x^{\,k} \;=\; n(n-1)\,x^{2}\,(1+x)^{\,n-2}.$$

Our goal involves $$k^{2}\,{}^{20}C_{k}$$, which we split as

$$k^{2} = k(k-1) + k.$$

Therefore

$$\sum_{k=0}^{20} k^{2}\,{}^{20}C_{k} \;=\;\sum_{k=0}^{20} k(k-1)\,{}^{20}C_{k} \;+\;\sum_{k=0}^{20} k\,{}^{20}C_{k}.$$

We now evaluate each part at $$x=1$$, because setting $$x=1$$ in the derived identities converts every factor $$(1+x)$$ into $$2$$ and removes any residual $$x$$ powers.

First part:

Using the second identity with $$n=20$$ and $$x=1$$, we have

$$\sum_{k=0}^{20} k(k-1)\,{}^{20}C_{k} \;=\; 20\cdot19\cdot1^{2}\,(1+1)^{18} \;=\;20\cdot19\cdot2^{18}.$$

Second part:

Using the first identity with $$n=20$$ and $$x=1$$, we have

$$\sum_{k=0}^{20} k\,{}^{20}C_{k} \;=\;20\cdot1\,(1+1)^{19} \;=\;20\cdot2^{19}.$$

Adding the two evaluated sums gives

$$\sum_{k=0}^{20} k^{2}\,{}^{20}C_{k} \;=\; \bigl(20\cdot19\cdot2^{18}\bigr) \;+\;\bigl(20\cdot2^{19}\bigr).$$

We factor the common powers of $$2$$:

$$20\cdot19\cdot2^{18} + 20\cdot2^{19} \;=\;20\cdot2^{18}\bigl(19 + 2\bigr) \;=\;20\cdot2^{18}\cdot21.$$

Multiplying the integers,

$$20\times21 = 420,$$

so

$$S \;=\;420\,2^{18}.$$

The general form requested in the question is $$A(2^{\beta})$$. By direct comparison, we identify

$$A = 420,\quad \beta = 18.$$

Hence, the correct answer is Option B.

We begin with the expression

$$S \;=\;\sum_{i=1}^{20}\left(\dfrac{{}^{20}C_{\,i-1}}{{}^{20}C_{\,i}+{}^{20}C_{\,i-1}}\right)^{\!3}$$

For each index $$i$$ we isolate the inner fraction

$$T_i \;=\;\dfrac{{}^{20}C_{\,i-1}}{{}^{20}C_{\,i}+{}^{20}C_{\,i-1}}.$$

To simplify it, we recall the relation connecting two successive binomial coefficients

$$^{20}C_i \;=\;^{20}C_{\,i-1}\;\times\;\dfrac{20-(i-1)}{i}\;=\;^{20}C_{\,i-1}\;\times\;\dfrac{21-i}{i}.$$

Substituting this value in the denominator of $$T_i$$, we get

$$^{20}C_{\,i}+{}^{20}C_{\,i-1} \;=\;^{20}C_{\,i-1}\left(\dfrac{21-i}{i}+1\right) \;=\;^{20}C_{\,i-1}\left(\dfrac{21-i+i}{i}\right) \;=\;^{20}C_{\,i-1}\left(\dfrac{21}{i}\right).$$

Hence

$$T_i \;=\;\dfrac{^{20}C_{\,i-1}}{^{20}C_{\,i-1}\left(\dfrac{21}{i}\right)} \;=\;\dfrac{1}{\dfrac{21}{i}} \;=\;\dfrac{i}{21}.$$

We need the cube of this term, so

$$T_i^{\,3}\;=\;\left(\dfrac{i}{21}\right)^{\!3}\;=\;\dfrac{i^{3}}{21^{3}}.$$

Substituting back into the summation, we obtain

$$S \;=\;\sum_{i=1}^{20}\dfrac{i^{3}}{21^{3}} \;=\;\dfrac{1}{21^{3}}\sum_{i=1}^{20} i^{3}.$$

Now we state the standard formula for the sum of cubes:

$$\sum_{i=1}^{n} i^{3} \;=\;\left[\dfrac{n(n+1)}{2}\right]^{2}.$$

Putting $$n=20$$, we have

$$\sum_{i=1}^{20} i^{3} \;=\;\left[\dfrac{20\times 21}{2}\right]^{2} \;=\;(10\times 21)^{2} \;=\;210^{2} \;=\;44100.$$

Therefore

$$S \;=\;\dfrac{1}{21^{3}}\times 44100.$$

Since $$21^{3}=21\times 21^{2}=21\times 441=9261$$, we get

$$S \;=\;\dfrac{44100}{9261}.$$

Factorising the numerator, note that $$44100=441\times 100$$, and because $$9261=21\times 441$$, we can write

$$S \;=\;\dfrac{441\times 100}{21\times 441} \;=\;\dfrac{100}{21}.$$

The question states $$S=\dfrac{k}{21}$$, so by equating we find

$$k \;=\;100.$$

Hence, the correct answer is Option B.

The expansion of $$(a+b)^n$$ has its general, that is, $$(r+1)^{\text{th}}$$, term given by the binomial-theorem formula

$$T_{r+1}=\,^nC_r\,a^{\,n-r}\,b^{\,r}\,.$$

In the present problem we have

$$n=6,\qquad a=\dfrac{2}{x},\qquad b=x^{\log_8 x}.$$

The fourth term corresponds to $$r+1=4\; \Longrightarrow\; r=3$$, so according to the formula we write

$$T_4={}^{6}C_{3}\left(\dfrac{2}{x}\right)^{6-3}\left(x^{\log_8 x}\right)^{3}.$$

Calculating the binomial coefficient first,

$$^{6}C_{3}=\dfrac{6!}{3!\,3!}=20.$$

Now evaluate each power separately.

$$\left(\dfrac{2}{x}\right)^{6-3}=\left(\dfrac{2}{x}\right)^{3}=\dfrac{2^{3}}{x^{3}}=\dfrac{8}{x^{3}}.$$

$$\left(x^{\log_8 x}\right)^{3}=x^{3\log_8 x}\; \bigl(\text{because }(x^{m})^{k}=x^{mk}\bigr).$$

Multiplying all the factors we obtain

$$T_4=20\;\times\;\dfrac{8}{x^{3}}\;\times\;x^{3\log_8 x}=160\,x^{\,3\log_8 x-3}.$$

The statement of the question tells us that this fourth term is equal to $$20\times 8^{7}$$, so we set

$$160\,x^{\,3\log_8 x-3}=20\,8^{7}.$$

Dividing both sides by $$20$$ gives

$$8\,x^{\,3\log_8 x-3}=8^{7}.$$

Now divide both sides by $$8$$:

$$x^{\,3\log_8 x-3}=8^{6}.$$

The base $$8$$ is convenient, so let us represent $$x$$ itself as a power of $$8$$. Put

$$x=8^{k}\qquad(k\in\mathbb{R}).$$

Then

$$\log_8 x=\log_8\!\bigl(8^{k}\bigr)=k.$$

Substituting these into the exponent of $$x$$ in the previous equation, we get

$$x^{\,3\log_8 x-3}=x^{\,3k-3}=\bigl(8^{k}\bigr)^{3k-3}=8^{\,k\,(3k-3)}.$$

Therefore the equality becomes

$$8^{\,k\,(3k-3)}=8^{6}.$$

Since the bases are identical and positive (both equal to $$8$$), the exponents must be equal:

$$k\,(3k-3)=6.$$

Simplify this quadratic equation:

$$3k^{2}-3k-6=0 \;\Longrightarrow\; k^{2}-k-2=0\quad(\text{dividing by }3).$$

Factorising,

$$(k-2)(k+1)=0\;\Longrightarrow\;k=2\quad\text{or}\quad k=-1.$$

Each possible value of $$k$$ gives a corresponding value of $$x$$ via $$x=8^{k}$$:

$$k=2\;\Rightarrow\;x=8^{2}=64,$$

$$k=-1\;\Rightarrow\;x=8^{-1}=\dfrac{1}{8}.$$

Among the answer choices provided, only $$x=8^{2}$$ appears. Hence, the correct answer is Option D.

We begin with the expression

$$ (x + 10)^{50} + (x - 10)^{50}. $$

To expand a power of a binomial we recall the Binomial Theorem, stated as

$$ (a + b)^n \;=\; \sum_{k=0}^{n} \binom{n}{k}\, a^{\,n-k}\, b^{\,k}. $$

First we apply the theorem to each of the two terms separately, taking $$a = x$$ and $$b = 10$$ for the first term and $$b = -10$$ for the second:

$$ (x + 10)^{50} \;=\; \sum_{k=0}^{50} \binom{50}{k}\, x^{50-k}\, 10^{\,k}, $$

$$ (x - 10)^{50} \;=\; \sum_{k=0}^{50} \binom{50}{k}\, x^{50-k}\, (-10)^{\,k}. $$

Adding the two series we obtain

$$ (x + 10)^{50} + (x - 10)^{50} \;=\; \sum_{k=0}^{50} \binom{50}{k}\, x^{50-k}\,\bigl(10^{\,k} + (-10)^{\,k}\bigr). $$

Notice that when $$k$$ is odd, the two quantities $$10^{\,k}$$ and $$(-10)^{\,k}$$ have opposite signs, so their sum is zero. When $$k$$ is even, the signs are the same and their sum is simply twice the positive value. Therefore only the even values of $$k$$ survive, each with an extra factor of $$2$$:

$$ (x + 10)^{50} + (x - 10)^{50} \;=\; \sum_{\substack{k = 0 \\ k \text{ even}}}^{50} 2\,\binom{50}{k}\, x^{50-k}\, 10^{\,k}. $$

We compare this with the required form

$$ a_0 + a_1 x + a_2 x^2 + \dots + a_{50} x^{50}, $$

and identify each coefficient $$a_r$$ with the term in which the power of $$x$$ is $$r$$. Remember that the power of $$x$$ is $$50 - k$$. We now compute the two specific coefficients we need.

Constant term $$a_0$$: For the constant term the power of $$x$$ must be $$0$$. Setting $$50 - k = 0$$ gives $$k = 50$$, which is indeed even. Substituting $$k = 50$$ in the general term, we obtain

$$ a_0 \;=\; 2 \,\binom{50}{50}\,10^{50}. $$

Since $$\binom{50}{50} = 1$$, this simplifies to

$$ a_0 \;=\; 2 \times 10^{50}. $$

Coefficient $$a_2$$ of $$x^2$$: Here we need the power of $$x$$ to be $$2$$. Setting $$50 - k = 2$$ gives $$k = 48$$, which is also even. Substituting $$k = 48$$ gives

$$ a_2 \;=\; 2 \,\binom{50}{48}\,10^{48}. $$

Because of the symmetry of binomial coefficients, we have $$\binom{50}{48} = \binom{50}{2}$$. Using

$$ \binom{50}{2} \;=\; \frac{50 \times 49}{2} \;=\; 1225, $$

we obtain

$$ a_2 \;=\; 2 \times 1225 \times 10^{48}. $$

The desired ratio $$\dfrac{a_2}{a_0}$$:

$$ \frac{a_2}{a_0} \;=\; \frac{2 \times 1225 \times 10^{48}}{2 \times 10^{50}} \;=\; \frac{1225 \times 10^{48}}{10^{50}} \;=\; \frac{1225}{10^{2}} \;=\; \frac{1225}{100} \;=\; 12.25. $$

Hence, the correct answer is Option C.

We have to evaluate the expression

$$E(x)=\left(x+\sqrt{x^{3}-1}\right)^{6}+\left(x-\sqrt{x^{3}-1}\right)^{6},\qquad x\gt 1,$$

and then add the coefficients of all those terms in the expansion whose powers of $$x$$ are even.

First, recall the Binomial Theorem in the form $$\left(a+b\right)^{n}=\displaystyle\sum_{k=0}^{n}\binom{n}{k}a^{\,n-k}b^{\,k}.$$

Using this with $$a=x,\;b=\sqrt{x^{3}-1},\;n=6$$ gives

$$\left(x\pm\sqrt{x^{3}-1}\right)^{6}=\sum_{k=0}^{6}\binom{6}{k}\,x^{\,6-k}\left(\pm\sqrt{x^{3}-1}\right)^{k}.$$

Now we add the “+” and “−” versions. For odd $$k$$ the two terms cancel because one has a plus sign and the other a minus sign; for even $$k$$ they double because both signs become the same. Hence

$$E(x)=2\sum_{\substack{k=0\\k\text{ even}}}^{6}\binom{6}{k}\,x^{\,6-k}\Bigl(\sqrt{x^{3}-1}\Bigr)^{k}.$$

Put $$k=2m$$ (so $$m=0,1,2,3$$). We obtain

$$E(x)=2\sum_{m=0}^{3}\binom{6}{2m}\,x^{\,6-2m}\left(\sqrt{x^{3}-1}\right)^{2m}.$$

Because $$\left(\sqrt{x^{3}-1}\right)^{2m}=\left(x^{3}-1\right)^{m},$$ the expression becomes

$$E(x)=2\sum_{m=0}^{3}\binom{6}{2m}\,x^{\,6-2m}\left(x^{3}-1\right)^{m}.$$

Next, expand $$\left(x^{3}-1\right)^{m}$$ again by the Binomial Theorem:

$$\left(x^{3}-1\right)^{m}=\sum_{r=0}^{m}\binom{m}{r}(x^{3})^{r}(-1)^{\,m-r}=\sum_{r=0}^{m}\binom{m}{r}(-1)^{\,m-r}x^{\,3r}.$$

Substituting this inside $$E(x)$$ gives

$$E(x)=2\sum_{m=0}^{3}\binom{6}{2m}\sum_{r=0}^{m}\binom{m}{r}(-1)^{\,m-r}\,x^{\,6-2m+3r}.$$

Thus each term has coefficient

$$2\binom{6}{2m}\binom{m}{r}(-1)^{\,m-r}$$

and power of $$x$$ equal to $$n=6-2m+3r.$$ We now list all pairs $$(m,r)$$, compute $$n$$ and keep only those where $$n$$ is even.

• $$m=0$$: only $$r=0$$ is possible.
  Exponent $$n=6-0+0=6\;( \text{even}).$$
  Coefficient $$2\binom{6}{0}\binom{0}{0}(-1)^{0}=2.$$

• $$m=1$$: $$r=0,1.$$
  $$r=0$$: $$n=6-2+0=4\;(\text{even}),$$ coefficient $$2\binom{6}{2}\binom{1}{0}(-1)^{1}=2\cdot15\cdot1\cdot(-1)=-30.$$
  $$r=1$$: $$n=6-2+3=7\;(\text{odd}),$$ ignored.

• $$m=2$$: $$r=0,1,2.$$
  $$r=0$$: $$n=6-4+0=2\;(\text{even}),$$ coefficient $$2\binom{6}{4}\binom{2}{0}(-1)^{2}=2\cdot15\cdot1\cdot1=30.$$
  $$r=1$$: $$n=6-4+3=5\;(\text{odd}),$$ ignored.
  $$r=2$$: $$n=6-4+6=8\;(\text{even}),$$ coefficient $$2\binom{6}{4}\binom{2}{2}(-1)^{0}=2\cdot15\cdot1\cdot1=30.$$

• $$m=3$$: $$r=0,1,2,3.$$
  $$r=0$$: $$n=6-6+0=0\;(\text{even}),$$ coefficient $$2\binom{6}{6}\binom{3}{0}(-1)^{3}=2\cdot1\cdot1\cdot(-1)=-2.$$
  $$r=1$$: $$n=6-6+3=3\;(\text{odd}),$$ ignored.
  $$r=2$$: $$n=6-6+6=6\;(\text{even}),$$ coefficient $$2\binom{6}{6}\binom{3}{2}(-1)^{1}=2\cdot1\cdot3\cdot(-1)=-6.$$
  $$r=3$$: $$n=6-6+9=9\;(\text{odd}),$$ ignored.

We now add all coefficients corresponding to even exponents:

Constant term (degree 0): $$-2$$

Degree 2: $$+30$$

Degree 4: $$-30$$

Degree 6: $$2+(-6)=-4$$

Degree 8: $$+30$$

Sum of all these coefficients:

$$-2+30-30-4+30 = 24.$$

Hence, the required sum of coefficients of the even-degree terms is $$24$$.

Hence, the correct answer is Option C.

First we observe that the given expression is a binomial of the form $$(a+b)^n$$ with

$$a=\dfrac{x^{3}}{3}, \qquad b=\dfrac{3}{x}, \qquad n=8.$$

For any binomial expansion $$(a+b)^n,$$ the general term (also called the $$(k+1)$$-th term when we start counting from $$k=0$$) is

$$T_{k+1}= \binom{n}{k}\, a^{\,n-k}\, b^{\,k}.$$

The “middle term’’ is the term that comes exactly in the centre of the expansion. When the exponent $$n$$ is even, i.e. $$n=2m$$, there is a single middle term, namely the $$(m+1)$$-th term. Here $$n=8=2\times4$$, so $$m=4$$ and the middle term is the $$(4+1)$$-th term, that is $$T_{5}$$.

Putting $$k=4$$ in the general formula, we obtain

$$T_{5}= \binom{8}{4}\left(\dfrac{x^{3}}{3}\right)^{8-4}\left(\dfrac{3}{x}\right)^{4}.$$

Now we simplify each factor carefully.

First, $$\binom{8}{4}=70,$$ because $$\binom{8}{4}=\dfrac{8!}{4!\,4!}=70.$$

Next,

$$\left(\dfrac{x^{3}}{3}\right)^{8-4}=\left(\dfrac{x^{3}}{3}\right)^{4} =\dfrac{x^{3\times4}}{3^{4}} =\dfrac{x^{12}}{81}.$$

Also,

$$\left(\dfrac{3}{x}\right)^{4} =\dfrac{3^{4}}{x^{4}} =\dfrac{81}{x^{4}}.$$

Multiplying the two powers of $$x$$ together, we have

$$\dfrac{x^{12}}{81}\times\dfrac{81}{x^{4}} =x^{12}\times\dfrac{81}{81}\times\dfrac{1}{x^{4}} =x^{12}\times\dfrac{1}{x^{4}} =x^{8}.$$

Hence the middle term becomes

$$T_{5}=70\,x^{8}.$$

According to the question, this value equals $$5670$$, so we set

$$70\,x^{8}=5670.$$

Dividing both sides by $$70$$, we get

$$x^{8}=\dfrac{5670}{70}=81.$$

We recognize that $$81=3^{4},$$ so

$$x^{8}=3^{4}.$$

Taking the real eighth roots on both sides gives

$$x=\pm 3^{\,4/8}= \pm 3^{1/2}=\pm\sqrt{3}.$$

Thus there are two real solutions: $$x=\sqrt{3}\quad\text{and}\quad x=-\sqrt{3}.$$

Finally, the sum of these real values is

$$\sqrt{3}+(-\sqrt{3})=0.$$

Hence, the correct answer is Option A.

We have been told that the three binomial coefficients $$^nC_4,\; ^nC_5,\; ^nC_6$$ form an arithmetic progression. In an arithmetic progression the middle term is the average of its neighbours, so the defining condition is

$$2\,^nC_5 \;=\; ^nC_4 + ^nC_6.$$

First recall the factorial form of a binomial coefficient:

$$^nC_r \;=\; \dfrac{n!}{r!\,(n-r)!}.$$

There is also a very convenient ratio formula derived straight from the factorial definition:

$$^nC_{\,r+1} \;=\; \dfrac{n-r}{r+1}\; ^nC_r.$$ This will allow us to rewrite every term in the above equation in terms of the single quantity $$^nC_5$$.

To express $$^nC_4$$ in terms of $$^nC_5$$ we set $$r=4$$ in the ratio formula and obtain

$$^nC_5 \;=\; \dfrac{n-4}{5}\; ^nC_4 \;\;\Longrightarrow\;\; ^nC_4 \;=\; ^nC_5 \;\dfrac{5}{\,n-4}.$$

Next, to express $$^nC_6$$ in terms of $$^nC_5$$ we set $$r=5$$ in the same ratio formula and get

$$^nC_6 \;=\; \dfrac{n-5}{6}\; ^nC_5.$$

Now substitute these two expressions into the arithmetic-progression condition:

$$2\,^nC_5 \;=\; ^nC_5\,\dfrac{5}{n-4} \;+\; ^nC_5\,\dfrac{n-5}{6}.$$

Because $$^nC_5$$ is a common, non-zero factor, we may divide both sides by it:

$$2 \;=\; \dfrac{5}{n-4} \;+\; \dfrac{n-5}{6}.$$

To clear the denominators multiply every term by $$6\,(n-4):$$

$$2 \times 6\,(n-4) \;=\; 5 \times 6 \;+\; (n-5)\,(n-4).$$

Simplifying the left-hand side:

$$12\,(n-4).$$

Simplifying the right-hand side step by step gives

$$30 + (n-5)(n-4).$$

Expanding the product $$(n-5)(n-4)$$:

$$n^2 - 4n - 5n + 20 \;=\; n^2 - 9n + 20.$$

So the equation becomes

$$12(n-4) \;=\; 30 + n^2 - 9n + 20.$$

Combine the constants on the right:

$$12(n-4) \;=\; n^2 - 9n + 50.$$

Expand the left-hand side:

$$12n - 48 \;=\; n^2 - 9n + 50.$$

Now bring every term to the right to obtain a standard quadratic equation:

$$0 \;=\; n^2 - 9n + 50 - 12n + 48,$$

which simplifies to

$$n^2 - 21n + 98 \;=\; 0.$$

We solve this quadratic by the discriminant method. The discriminant is

$$\Delta \;=\; (-21)^2 - 4 \times 1 \times 98 \;=\; 441 - 392 \;=\; 49.$$

Since $$\sqrt{\Delta}=7,$$ the roots are

$$n \;=\; \dfrac{21 \pm 7}{2}.$$

This gives the two numerical values

$$n_1 = \dfrac{21 + 7}{2} = \dfrac{28}{2} = 14,$$ $$n_2 = \dfrac{21 - 7}{2} = \dfrac{14}{2} = 7.$$

Both numbers are indeed > 6, so both satisfy the algebraic condition. However, among the options offered (9, 14, 12, 11) only $$14$$ appears. Consequently the permissible value of $$n$$ from the given list is $$14$$.

Hence, the correct answer is Option B.

We consider the binomial expansion $$ (x+1)^n = \sum_{k=0}^{n} \binom{n}{k} x^{\,k}. $$

The coefficient of $$x^{\,k}$$ is $$\binom{n}{k}.$$ Three consecutive coefficients will therefore be $$ \binom{n}{r-1}, \; \binom{n}{r}, \; \binom{n}{r+1} $$ for some integer $$r$$ with $$1\le r\le n-1.$$

According to the question, these three coefficients are in the ratio $$2:15:70.$$ So we can write

$$ \frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{15}{2}, \qquad \frac{\binom{n}{r+1}}{\binom{n}{r}}=\frac{70}{15}=\frac{14}{3}. $$

First we recall the standard identity connecting consecutive binomial coefficients:

Formula: $$ \frac{\binom{n}{k}}{\binom{n}{k-1}}=\frac{n-k+1}{k}. $$

Applying this to the first ratio with $$k=r,$$ we get

$$ \frac{\binom{n}{r}}{\binom{n}{r-1}} =\frac{n-r+1}{r}. $$

Set this equal to $$15/2$$ as required by the given ratio:

$$ \frac{n-r+1}{r}=\frac{15}{2}. $$

Cross-multiplying,

$$ 2(n-r+1)=15r, $$

$$ 2n-2r+2=15r, $$

$$ 2n = 17r-2, $$

$$ n = \frac{17r-2}{2}. $$

For $$n$$ to be an integer, $$17r-2$$ must be even. Because 17 is odd, $$r$$ itself must be even. Let us put $$r=2k$$ where $$k$$ is a positive integer.

Substituting $$r=2k$$ in the expression for $$n$$ we obtain

$$ n = \frac{17(2k)-2}{2}=\frac{34k-2}{2}=17k-1. $$

Now we turn to the second ratio. Using the same identity with $$k=r+1,$$ we have

$$ \frac{\binom{n}{r+1}}{\binom{n}{r}} =\frac{n-(r+1)+1}{r+1} =\frac{n-r}{r+1}. $$

This must equal $$\dfrac{14}{3}.$$ Therefore

$$ \frac{n-r}{r+1}=\frac{14}{3}. $$

Substituting the earlier expressions $$n=17k-1$$ and $$r=2k,$$ we get

$$ \frac{(17k-1)-2k}{2k+1}=\frac{14}{3}, $$

$$ \frac{15k-1}{2k+1}=\frac{14}{3}. $$

Cross-multiplying once more,

$$ 3(15k-1)=14(2k+1), $$

$$ 45k-3=28k+14, $$

$$ 45k-28k=14+3, $$

$$ 17k = 17, $$

$$ k = 1. $$

Putting $$k=1$$ back, we obtain

$$ r = 2k = 2, \qquad n = 17k-1 = 16. $$

Thus the three required coefficients are

$$ \binom{16}{1},\; \binom{16}{2},\; \binom{16}{3}. $$

We evaluate them one by one:

$$ \binom{16}{1}=16, $$

$$ \binom{16}{2}=\frac{16\cdot15}{2}=120, $$

$$ \binom{16}{3}=\frac{16\cdot15\cdot14}{3\cdot2\cdot1}=560. $$

The three numbers $$16,\,120,\,560$$ are indeed in the ratio $$2:15:70$$ (dividing each by 8 confirms this).

The average (arithmetic mean) of these three coefficients is

$$ \text{Average}=\frac{16+120+560}{3}=\frac{696}{3}=232. $$

Hence, the correct answer is Option D.

We begin with the given expression

$$\bigl(1 + ax + bx^2\bigr)\,(1 - 3x)^{15}$$

and are told that, when it is expanded in powers of $$x$$, the coefficients of $$x^2$$ and $$x^3$$ are both equal to zero. Our task is to determine the real numbers $$a$$ and $$b$$ satisfying those two conditions.

The expansion of $$(1 - 3x)^{15}$$ is obtained through the Binomial Theorem, which states

$$\bigl(1 + t\bigr)^n \;=\; \sum_{k = 0}^{n} \binom{n}{k}\, t^{\,k}.$$

Here we have $$t = -3x$$ and $$n = 15$$, so we write

$$\bigl(1 - 3x\bigr)^{15} \;=\; \sum_{k = 0}^{15} \binom{15}{k}\,(-3x)^{k} \;=\; \sum_{k = 0}^{15} \binom{15}{k}\,(-3)^{k}\,x^{k}.$$

Thus, for each power $$k$$, the coefficient of $$x^k$$ in $$(1 - 3x)^{15}$$ is

$$\binom{15}{k}\,(-3)^{k}.$$

Next, we multiply this series by $$1 + ax + bx^2$$. The coefficient of any specific power of $$x$$ in the product comes from matching powers whose exponents add up to that specific power. We will impose the conditions for $$x^2$$ and $$x^3$$ in turn.

Coefficient of $$x^2$$

The power $$x^2$$ in the product can arise in exactly three ways:

(1) The factor $$1$$ (from $$1 + ax + bx^2$$) times the $$x^2$$ term of $$(1 - 3x)^{15}$$;
(2) The factor $$ax$$ times the $$x^1$$ term of $$(1 - 3x)^{15}$$;
(3) The factor $$bx^2$$ times the constant term (the $$x^0$$ term) of $$(1 - 3x)^{15}$$.

We compute each required coefficient directly:

$$\text{Coeff}_{x^2}\bigl((1 - 3x)^{15}\bigr) = \binom{15}{2}(-3)^2,$$
$$\text{Coeff}_{x^1}\bigl((1 - 3x)^{15}\bigr) = \binom{15}{1}(-3)^1,$$
$$\text{Coeff}_{x^0}\bigl((1 - 3x)^{15}\bigr) = \binom{15}{0}(-3)^0.$$

Evaluating the numerical values:

$$\binom{15}{2} = 105,\quad (-3)^2 = 9 \;\Longrightarrow\; 105 \times 9 = 945,$$
$$\binom{15}{1} = 15,\quad (-3)^1 = -3 \;\Longrightarrow\; 15 \times (-3) = -45,$$
$$\binom{15}{0} = 1,\quad (-3)^0 = 1 \;\Longrightarrow\; 1 \times 1 = 1.$$

Hence the coefficient of $$x^2$$ in the entire product is

$$945 \;+\; a(-45) \;+\; b(1).$$

Because this coefficient is required to be zero, we obtain our first linear equation:

$$945 \;-\; 45a \;+\; b = 0.$$

Solving for $$b$$ gives

$$b = 45a - 945.$$

Coefficient of $$x^3$$

In an analogous manner, the power $$x^3$$ can arise from:

(1) $$1$$ multiplying the $$x^3$$ term of $$(1 - 3x)^{15}$$;
(2) $$ax$$ multiplying the $$x^2$$ term of $$(1 - 3x)^{15}$$;
(3) $$bx^2$$ multiplying the $$x^1$$ term of $$(1 - 3x)^{15}$$.

We already know the $$x^2$$ and $$x^1$$ coefficients from above; we now find the $$x^3$$ coefficient:

$$\text{Coeff}_{x^3}\bigl((1 - 3x)^{15}\bigr) = \binom{15}{3}(-3)^3.$$

Compute this value carefully:

$$\binom{15}{3} = 455,\quad (-3)^3 = -27 \;\Longrightarrow\; 455 \times (-27) = -12\,285.$$

Therefore the overall $$x^3$$ coefficient in the product equals

$$-12\,285 \;+\; a \cdot 945 \;+\; b \cdot (-45).$$

This must also be zero, giving the second equation:

$$-12\,285 \;+\; 945a \;-\; 45b = 0.$$

Substituting $$b = 45a - 945$$ into the second equation

We write

$$-12\,285 \;+\; 945a \;-\; 45(45a - 945) = 0.$$

Distribute the factor $$-45$$ inside the parentheses:

$$-12\,285 \;+\; 945a \;-\; 45 \cdot 45a \;+\; 45 \cdot 945 = 0.$$

Since $$45 \cdot 45a = 2\,025a$$ and $$45 \cdot 945 = 42\,525$$, the equation becomes

$$-12\,285 \;+\; 945a \;-\; 2\,025a \;+\; 42\,525 = 0.$$

Combine the like terms:

$$(-12\,285 + 42\,525) \;+\; (945a - 2\,025a) = 0,$$
$$30\,240 \;-\; 1\,080a = 0.$$

Rearranging gives

$$-1\,080a = -30\,240 \;\Longrightarrow\; a = \frac{30\,240}{1\,080} = 28.$$

Finding $$b$$ from $$b = 45a - 945$$

Substituting $$a = 28$$:

$$b = 45 \times 28 - 945.$$

Calculate step by step:

$$45 \times 28 = 1\,260,$$
$$1\,260 - 945 = 315.$$

We have now obtained

$$a = 28, \qquad b = 315.$$

Therefore the ordered pair $$\bigl(a,\,b\bigr)$$ equals $$\bigl(28,\,315\bigr).$$

Examining the provided options, this matches Option A.

Hence, the correct answer is Option A.

We begin with the given binomial expression

$$\left(\sqrt{x^{\frac{1}{1+\log_{10}x}}}+x^{\frac{1}{12}}\right)^{6}.$$

Write the two terms inside the bracket in a simpler exponential form.

The first term is a square root. Using the law $$\sqrt{A}=A^{1/2},$$ we have

$$\sqrt{x^{\frac{1}{1+\log_{10}x}}}= \left(x^{\frac{1}{1+\log_{10}x}}\right)^{\frac12}=x^{\frac{1}{2(1+\log_{10}x)}}.$$

The second term is already a single power:

$$x^{\frac{1}{12}}.$$

So let us set

$$a = x^{\frac{1}{2(1+\log_{10}x)}}, \qquad b = x^{\frac{1}{12}}.$$

We have the overall power $$\;6,$$ so the binomial theorem gives the general term

$$T_{r+1}= \binom{6}{r}\,a^{\,6-r}\,b^{\,r}\qquad(r=0,1,2,\dots ,6).$$

The fourth term corresponds to $$r=3,$$ because the counting starts from $$r=0.$$ Therefore

$$T_4 = \binom{6}{3}\,a^{\,6-3}\,b^{\,3}=\binom{6}{3}\,a^{3}\,b^{3}.$$

Since $$\binom{6}{3}=20,$$ this becomes

$$T_4 = 20\,a^{3}\,b^{3}.$$

According to the statement of the problem, this fourth term equals 200, i.e.

$$20\,a^{3}\,b^{3}=200.$$

Dividing by 20 gives

$$a^{3}\,b^{3}=10.$$

Taking the cube root on both sides,

$$(ab)=10^{1/3}.$$

Now substitute the explicit expressions for $$a$$ and $$b$$:

$$ab = x^{\frac{1}{2(1+\log_{10}x)}}\;x^{\frac{1}{12}} = x^{\;\frac{1}{2(1+\log_{10}x)}+\frac{1}{12}} = 10^{1/3}.$$

To eliminate the logarithm, set $$x=10^{k}$$ with $$k>0$$ (because $$x>1$$). Then $$\log_{10}x=k.$$

Compute the combined exponent of $$x$$ (or equivalently of $$10^{k}$$). First note

$$\frac{1}{2(1+\log_{10}x)}=\frac{1}{2(1+k)}.$$

Hence the total exponent in the product $$ab$$ is

$$\frac{1}{2(1+k)}+\frac{1}{12}.$$

Put these over a common denominator:

$$\frac{1}{2(1+k)}+\frac{1}{12}= \frac{6}{12(1+k)}+\frac{1+k}{12(1+k)} =\frac{6+1+k}{12(1+k)} =\frac{k+7}{12(1+k)}.$$

Therefore

$$ab = x^{\frac{k+7}{12(1+k)}}=(10^{k})^{\frac{k+7}{12(1+k)}} =10^{\,k\,\frac{k+7}{12(1+k)}}.$$

The equality $$ab=10^{1/3}$$ now reads

$$10^{\,k\,\frac{k+7}{12(1+k)}}=10^{1/3}.$$

Because the bases are equal (both are 10), we set the exponents equal:

$$k\,\frac{k+7}{12(1+k)}=\frac13.$$

Multiply both sides by $$12(1+k)$$ to clear the denominator:

$$k(k+7)=4(1+k).$$

Expand both sides:

$$k^{2}+7k=4k+4.$$

Bring all terms to one side:

$$k^{2}+7k-4k-4=0 \;\;\Longrightarrow\;\; k^{2}+3k-4=0.$$

This is a quadratic equation. Using the quadratic formula $$k=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ with $$a=1,\;b=3,\;c=-4,$$ we get

$$k=\frac{-3\pm\sqrt{3^{2}-4(1)(-4)}}{2} =\frac{-3\pm\sqrt{9+16}}{2} =\frac{-3\pm5}{2}.$$

The two roots are

$$k=\frac{2}{2}=1,\qquad k=\frac{-8}{2}=-4.$$

Since $$k>0$$ (because $$x>1$$), we take $$k=1.$$ Thus

$$x=10^{k}=10^{1}=10.$$

Hence, the correct answer is Option D.

We recall the binomial theorem, which says that in the expansion of $$(a+b)^n$$ the general term (often called the $$(r+1)^{\text{th}}$$ term, starting the count from $$r=0$$) is

$$T_{r+1}=\,\,{^{n}C_{r}}\,a^{\,n-r}\,b^{\,r}.$$

In the given expression $$(1+x^{\log_{2}x})^{5}$$ we have $$a=1,\;b=x^{\log_{2}x},\;n=5.$$

The third term corresponds to $$r=2$$ (because for the first term $$r=0$$, for the second $$r=1$$, and for the third $$r=2$$). Substituting these values in the formula we obtain

$$T_{3}=\,\,{^{5}C_{2}}\,(1)^{5-2}\,\bigl(x^{\log_{2}x}\bigr)^{2}.$$

Since any power of 1 is 1, the factor $$(1)^{5-2}$$ is simply 1, and we are left with

$$T_{3}=\,\,{^{5}C_{2}}\;x^{2\log_{2}x}.$$

We know that $${^{5}C_{2}}=\dfrac{5!}{2!\,3!}=10,$$ so

$$T_{3}=10\,x^{2\log_{2}x}.$$

The problem states that this third term equals 2560, so we set

$$10\,x^{2\log_{2}x}=2560.$$

Dividing both sides by 10 gives

$$x^{2\log_{2}x}=256.$$

Next we express 256 as a power of 2. We know $$256=2^{8},$$ so the equation becomes

$$x^{2\log_{2}x}=2^{8}.$$

To simplify the left-hand side, let us write $$\log_{2}x=k.$$ Then $$x=2^{k}.$$ Substituting this into the left-hand side gives

$$x^{2\log_{2}x}=(2^{k})^{2k}=2^{k\cdot2k}=2^{2k^{2}}.$$

Therefore our equation is now

$$2^{2k^{2}}=2^{8}.$$

Because the bases are equal, we equate the exponents:

$$2k^{2}=8.$$

Dividing by 2 yields

$$k^{2}=4.$$

Taking square roots,

$$k=\pm2.$$

Recall that $$k=\log_{2}x,$$ so we have two cases:

1. $$\log_{2}x=2\;\;\Longrightarrow\;\;x=2^{2}=4.$$

2. $$\log_{2}x=-2\;\;\Longrightarrow\;\;x=2^{-2}=\dfrac{1}{4}.$$

Both values are positive and hence admissible, but only one of them appears among the given options. Inspecting the options, we see that $$\dfrac{1}{4}$$ is listed, whereas 4 is not.

Hence, the correct answer is Option 4.

We have to calculate the coefficient of $$x^{18}$$ in the product

$$ (1 + x)\,(1 - x)^{10}\,(1 + x + x^2)^9. $$

First, recall the useful identity

$$ 1 + x + x^2 \;=\; \frac{1 - x^3}{1 - x}. $$

Raising both sides to the 9-th power gives

$$ (1 + x + x^2)^9 \;=\; \left(\frac{1 - x^3}{1 - x}\right)^9 \;=\; (1 - x^3)^9\,(1 - x)^{-9}. $$

Substituting this into the original product, we get

$$ (1 + x)\,(1 - x)^{10}\,(1 + x + x^2)^9 \;=\; (1 + x)\,(1 - x)^{10}\,(1 - x^3)^9\,(1 - x)^{-9}. $$

Because the exponents on $$(1 - x)$$ add algebraically, we have

$$ (1 - x)^{10}\,(1 - x)^{-9} \;=\; (1 - x)^{10-9} \;=\; (1 - x)^1 \;=\; (1 - x). $$

So the whole product simplifies to

$$ (1 + x)\,(1 - x)\,(1 - x^3)^9. $$

Now multiply the first two linear factors:

$$ (1 + x)\,(1 - x) \;=\; 1 - x^2. $$

Therefore the expression whose coefficient we need is

$$ (1 - x^2)\,(1 - x^3)^9. $$

Next we expand $$(1 - x^3)^9$$ with the binomial theorem. The binomial theorem states

$$ (1 + y)^n \;=\; \sum_{r=0}^{n} \binom{n}{r}\,y^r. $$

With $$y = -x^3$$ and $$n = 9,$$ this gives

$$ (1 - x^3)^9 \;=\; \sum_{r=0}^{9} \binom{9}{r}\,(-1)^r\,x^{3r}. $$

Multiplying by $$1 - x^2,$$ we write

$$ (1 - x^2)\,(1 - x^3)^9 \;=\; (1)\,\bigl[(1 - x^3)^9\bigr] \;-\; x^2\,(1 - x^3)^9. $$

Substituting the series for $$(1 - x^3)^9$$ in each part, we obtain

$$ \begin{aligned} (1 - x^2)\,(1 - x^3)^9 &=\; \sum_{r=0}^{9} \binom{9}{r}(-1)^r x^{3r} \;-\; \sum_{r=0}^{9} \binom{9}{r}(-1)^r x^{3r+2}. \\ \end{aligned} $$

The two sums now clearly show every power of $$x$$ that can occur. To get the coefficient of $$x^{18},$$ we examine each sum separately.

• In the first sum, the exponent is $$3r.$$   Setting $$3r = 18$$ gives $$r = 6,$$ which is indeed an integer between 0 and 9, so there is a contribution.

• In the second sum, the exponent is $$3r + 2.$$   Setting $$3r + 2 = 18$$ gives $$3r = 16,$$ or $$r = \dfrac{16}{3},$$ which is not an integer.   Therefore no term in the second sum can produce $$x^{18}.$$

Thus the only contribution to the $$x^{18}$$ coefficient comes from the first sum with $$r = 6.$$ The corresponding term is

$$ \binom{9}{6}(-1)^6 x^{18}. $$

Since $$(-1)^6 = +1,$$ the coefficient is simply

$$ \binom{9}{6} \;=\; \binom{9}{3} \;=\; \frac{9\times8\times7}{3\times2\times1} \;=\; 84. $$

Hence, the correct answer is Option A.

We observe that every term of the given series has the general form $$\bigl(2+3k\bigr)\,{}^{20}C_k$$ where the index $$k$$ runs from $$0$$ to $$20$$. Indeed, for $$k=0$$ we get $$2\,{}^{20}C_0$$, for $$k=1$$ we get $$\bigl(2+3\bigr)\,{}^{20}C_1=5\,{}^{20}C_1$$, and so on up to $$k=20$$ which gives $$\bigl(2+3\cdot20\bigr)\,{}^{20}C_{20}=62\,{}^{20}C_{20}$$.

So we can write the required sum as

$$S=\sum_{k=0}^{20}\bigl(2+3k\bigr)\,{}^{20}C_k.$$

Now we split this into two separate sums:

$$S=2\sum_{k=0}^{20}{}^{20}C_k+3\sum_{k=0}^{20}k\,{}^{20}C_k.$$

First we evaluate $$\displaystyle\sum_{k=0}^{20}{}^{20}C_k$$. By the binomial theorem we know the identity

$$(1+1)^{20}=\sum_{k=0}^{20}{}^{20}C_k\,1^{20-k}\,1^{k}$$

which simplifies to

$$\sum_{k=0}^{20}{}^{20}C_k=2^{20}.$$

Next we need $$\displaystyle\sum_{k=0}^{20}k\,{}^{20}C_k$$. A standard combinatorial identity gives

$$\sum_{k=0}^{n}k\,{}^{n}C_k=n\,2^{\,n-1},$$

so for $$n=20$$ we obtain

$$\sum_{k=0}^{20}k\,{}^{20}C_k=20\cdot2^{19}.$$

Substituting these two results back into the expression for $$S$$, we get

$$S=2\cdot2^{20}+3\cdot\bigl(20\cdot2^{19}\bigr).$$

Simplifying the first product, we have $$2\cdot2^{20}=2^{21}.$$ For the second product, we first multiply the constants: $$3\times20=60,$$ giving $$60\cdot2^{19}.$$ Hence

$$S=2^{21}+60\,2^{19}.$$ Now we factor out the common power $$2^{19}$$:

$$S=2^{19}\bigl(2^{2}+60\bigr)=2^{19}\bigl(4+60\bigr)=2^{19}\times64.$$

Recognising that $$64=2^{6}$$, we conclude

$$S=2^{19}\times2^{6}=2^{25}.$$

Hence, the correct answer is Option B.

First, we observe that the given expression is a product of two factors:

$$\left(\frac1{60}-\frac{x^{8}}{81}\right)\left(2x^{2}-\frac{3}{x^{2}}\right)^{6}.$$

The term independent of $$x$$ (that is, the constant term) will come from appropriate selections inside this product so that all powers of $$x$$ cancel out.

We begin by expanding the second factor using the binomial theorem. The binomial theorem states:

$$(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}\,a^{\,n-k}\,b^{\,k}.$$

Here we identify $$a = 2x^{2}$$, $$b = -\dfrac{3}{x^{2}}$$ and $$n = 6$$. Hence the general term in the expansion of $$\left(2x^{2}-\dfrac{3}{x^{2}}\right)^{6}$$ is

$$T_{k}= \binom{6}{k}\left(2x^{2}\right)^{6-k}\left(-\frac{3}{x^{2}}\right)^{k}.$$ Simplifying the powers of $$x$$, we have

$$T_{k}= \binom{6}{k}\,2^{\,6-k}\,(-3)^{k}\,x^{\,2(6-k)}\,x^{-2k} = \binom{6}{k}\,2^{\,6-k}\,(-3)^{k}\,x^{\,12-4k}.$$

So $$T_{k}$$ contributes a power $$x^{\,12-4k}$$.

Now we multiply each term $$T_{k}$$ by the two terms in the first factor $$\left(\dfrac1{60} - \dfrac{x^{8}}{81}\right)$$ and see when the overall power of $$x$$ becomes zero.

Case 1: Choose $$\dfrac1{60}$$. The resulting power of $$x$$ is unchanged, so we need

$$12-4k = 0 \quad\Longrightarrow\quad 4k = 12 \quad\Longrightarrow\quad k = 3.$$

For $$k = 3$$ we compute the coefficient:

$$\binom{6}{3} = 20,\qquad 2^{\,6-3}=2^{3}=8,\qquad (-3)^{3}=-27.$$

Multiplying,

$$\binom{6}{3}\,2^{3}\,(-3)^{3}=20 \times 8 \times (-27)= -4320.$$

We still have the factor $$\dfrac1{60}$$, so the contribution to the constant term is

$$\frac1{60}\times(-4320)= -\frac{4320}{60}= -72.$$

Case 2: Choose $$-\dfrac{x^{8}}{81}$$. The extra factor $$x^{8}$$ changes the power, so we now need

$$12-4k + 8 = 0 \quad\Longrightarrow\quad 20-4k = 0 \quad\Longrightarrow\quad k = 5.$$

For $$k = 5$$ we compute the coefficient:

$$\binom{6}{5}=6,\qquad 2^{\,6-5}=2,\qquad (-3)^{5} = -243.$$

Thus,

$$\binom{6}{5}\,2\,(-3)^{5}=6 \times 2 \times (-243)= -2916.$$

Multiplying by the prefactor $$-\dfrac1{81}$$ gives

$$-\frac1{81}\times(-2916)=\frac{2916}{81}=36.$$

Adding the two contributions obtained in Case 1 and Case 2, the net constant term is

$$-72 + 36 = -36.$$

Hence, the correct answer is Option D.

We begin with the expression

$$S(r)= {}^{20}C_r \cdot {}^{20}C_0 + {}^{20}C_{r-1} \cdot {}^{20}C_1 + {}^{20}C_{r-2} \cdot {}^{20}C_2 + \ldots + {}^{20}C_0 \cdot {}^{20}C_r.$$

The general term of this finite sum can be written as $$^{20}C_{r-k}\,^{20}C_k$$ where the index $$k$$ goes from $$0$$ up to $$r$$. Hence we may rewrite the whole sum compactly as

$$S(r)=\sum_{k=0}^{r} {}^{20}C_{r-k}\,{}^{20}C_k.$$

Now we recall (and explicitly state) a standard identity that comes directly from the Binomial Theorem:

$$\bigl(1+x\bigr)^{m}\bigl(1+x\bigr)^{n}=(1+x)^{\,m+n}.$$

When we expand each factor separately and then multiply, the coefficient of $$x^t$$ on the left-hand side is obtained by convolution of the two individual coefficient sequences. In symbols,

$$\bigl[(1+x)^m\bigr]\bigl[(1+x)^n\bigr]=\sum_{t=0}^{m+n}\Bigl[\sum_{k=0}^{t}{}^{m}C_k\,{}^{n}C_{t-k}\Bigr]x^{\,t}.$$

Comparing this with the right-hand side $$ (1+x)^{m+n} $$ we see that

$$\sum_{k=0}^{t}{}^{m}C_k\,{}^{n}C_{t-k}={} ^{\,m+n}C_t. \quad -(★)$$

This is the convolution identity for binomial coefficients.

In our problem we have $$m=20,\; n=20$$ and $$t=r$$. Therefore, by direct substitution into identity (★) we get

$$S(r)=\sum_{k=0}^{r}{}^{20}C_k\,{}^{20}C_{r-k}={} ^{\,40}C_r.$$

So the given sum is precisely the single binomial coefficient $$^{40}C_r$$.

Next we determine for which integer $$r$$ (restricted here to $$0\le r\le 20$$ because $$^{20}C_{r-k}$$ vanishes whenever the upper index $$20$$ is smaller than the lower one) the value of $$^{40}C_r$$ is largest.

We use another well-known fact about binomial coefficients:

For a fixed positive integer $$N$$, the sequence $$^{N}C_0,\,^{N}C_1,\,^{N}C_2,\ldots,\,^{N}C_N$$ increases strictly up to $$r=\bigl\lfloor N/2\bigr\rfloor$$ and then decreases symmetrically, with

$$^{N}C_r=^{N}C_{N-r}.$$

Here $$N=40$$ is even, so the unique largest term is achieved at exactly

$$r=\dfrac{N}{2}=20.$$

Since all permissible $$r$$ lie between $$0$$ and $$20$$ (inclusive), the maximum clearly occurs at this central value itself.

Hence, the correct answer is Option B.

We have the binomial $$\left(2^{1/3}+\dfrac{1}{2\,(3)^{1/3}}\right)^{10}$$. Denote

$$a = 2^{1/3}, \qquad b = \dfrac{1}{2\,(3)^{1/3}}.$$

The general term in the expansion of $$(a+b)^{n}$$ is, by the Binomial Theorem, stated first as

$$T_{r+1}=^nC_r\,a^{\,n-r}\,b^{\,r},$$

where $$r=0,1,2,\dots ,n$$. Here $$n=10$$.

Because counting starts with $$r=0$$ for the first term, the 5th term from the beginning corresponds to $$r=4$$. Substituting $$n=10$$ and $$r=4$$ gives

$$T_{5\text{ (beg.)}} =^{10}C_4\,a^{\,10-4}\,b^{\,4}=^{10}C_4\,a^{6}\,b^{4}.$$

For the 5th term from the end we note that there are $$n+1=11$$ terms in total, so the 5th from the end is the $$11-5+1=7$$-th term from the beginning; thus it has index $$r=6$$. Substituting $$r=6$$ yields

$$T_{5\text{ (end)}} =^{10}C_6\,a^{\,10-6}\,b^{\,6}=^{10}C_6\,a^{4}\,b^{6}.$$

Now we form the required ratio:

$$\dfrac{T_{5\text{ (beg.)}}}{T_{5\text{ (end)}}}= \dfrac{^{10}C_4\,a^{6}\,b^{4}}{^{10}C_6\,a^{4}\,b^{6}}.$$

Because $$^{10}C_4 = ^{10}C_6 = 210,$$ the binomial coefficients cancel, giving

$$\dfrac{T_{5\text{ (beg.)}}}{T_{5\text{ (end)}}}= \dfrac{a^{6}\,b^{4}}{a^{4}\,b^{6}} = \dfrac{a^{2}}{b^{2}} = \left(\dfrac{a}{b}\right)^{2}.$$

So we must first evaluate $$\dfrac{a}{b}$$. Substituting $$a=2^{1/3}$$ and $$b=\dfrac{1}{2\,(3)^{1/3}}$$, we get

$$\dfrac{a}{b}=2^{1/3}\;\div\;\dfrac{1}{2\,(3)^{1/3}} = 2^{1/3}\times 2\,(3)^{1/3}=2\cdot 2^{1/3}\,(3)^{1/3}.$$

Observing that $$2^{1/3}(3)^{1/3}=(2\cdot 3)^{1/3}=(6)^{1/3},$$ we simplify further:

$$\dfrac{a}{b}= 2\,(6)^{1/3}.$$

Therefore

$$\left(\dfrac{a}{b}\right)^{2} = \left(2\,(6)^{1/3}\right)^{2}=4\,(6)^{2/3}.$$

Because $$(6)^{2/3}=\left(6^{2}\right)^{1/3}=36^{1/3},$$ we may write

$$\left(\dfrac{a}{b}\right)^{2}=4\,(36)^{1/3}.$$

Thus the ratio of the 5th term from the beginning to the 5th term from the end is

$$4\,(36)^{1/3}:1.$$

Hence, the correct answer is Option B.

We have to find the coefficient of $$t^4$$ in the expression $$\left(\dfrac{1-t^{6}}{1-t}\right)^3$$.

First, recall the geometric‐series identity

$$\frac{1-x^{n}}{1-x}=1+x+x^{2}+x^{3}+\dots +x^{n-1}.$$

Applying this with $$x=t$$ and $$n=6$$, we get

$$\frac{1-t^{6}}{1-t}=1+t+t^{2}+t^{3}+t^{4}+t^{5}.$$

So our given expression becomes

$$\left(1+t+t^{2}+t^{3}+t^{4}+t^{5}\right)^3.$$

In this product each factor contributes a power of $$t$$ from the set $$\{0,1,2,3,4,5\}$$. To obtain the term $$t^4$$, we must select three exponents (one from each factor) whose sum is exactly $$4$$.

Let those exponents be $$i$$, $$j$$ and $$k$$. Then

$$i+j+k=4,$$

with the restrictions $$0\le i\le5,\;0\le j\le5,\;0\le k\le5.$$ Because the required sum is only $$4$$, every solution automatically satisfies $$i,j,k\le5$$, so we may ignore these upper limits.

The problem of counting ordered triples $$(i,j,k)$$ of non-negative integers satisfying $$i+j+k=4$$ is a standard stars-and-bars problem. The formula is:

Number of non-negative integer solutions of $$x_1+x_2+\dots+x_r = n$$ is $$\binom{n+r-1}{r-1}.$$

Here $$r=3$$ and $$n=4$$, so we have

$$\binom{4+3-1}{3-1}=\binom{6}{2}=15.$$

Each such ordered triple contributes $$1\cdot1\cdot1=1$$ to the coefficient because every individual term in the parenthesis has coefficient $$1$$. Therefore, summing over the $$15$$ different triples, the coefficient of $$t^4$$ is $$15$$.

Hence, the correct answer is Option C.

We need to find the smallest natural number $$n$$ such that the coefficient of $$x$$ in the expansion of $$\left(x^2 + \frac{1}{x^3}\right)^n$$ is $$^nC_{23}$$.

The general term in the binomial expansion of $$\left(x^2 + \frac{1}{x^3}\right)^n$$ is:

$$T_{r+1} = \binom{n}{r} (x^2)^{n-r} \left(\frac{1}{x^3}\right)^r = \binom{n}{r} x^{2(n-r)} \cdot x^{-3r} = \binom{n}{r} x^{2n - 2r - 3r} = \binom{n}{r} x^{2n - 5r}$$

For the coefficient of $$x$$, we need the power of $$x$$ to be 1:

$$2n - 5r = 1$$ $$-(1)$$

This gives $$r = \frac{2n-1}{5}$$. For $$r$$ to be a non-negative integer, $$2n - 1$$ must be divisible by 5.

The coefficient of $$x$$ in the expansion is $$\binom{n}{r}$$, and we are told this equals $$\binom{n}{23}$$:

$$\binom{n}{r} = \binom{n}{23}$$

Using the property of binomial coefficients, $$\binom{n}{r} = \binom{n}{23}$$ implies either:

Case 1: $$r = 23$$, or

Case 2: $$r = n - 23$$ (using the symmetry property $$\binom{n}{k} = \binom{n}{n-k}$$)

Case 1: If $$r = 23$$, substituting into equation $$(1)$$:

$$2n - 5(23) = 1$$

$$2n = 116$$

$$n = 58$$

Case 2: If $$r = n - 23$$, substituting into equation $$(1)$$:

$$2n - 5(n-23) = 1$$

$$2n - 5n + 115 = 1$$

$$-3n = -114$$

$$n = 38$$

We need to verify that $$r$$ is a valid non-negative integer in each case:

For $$n = 38$$: $$r = 38 - 23 = 15$$, and $$0 \leq 15 \leq 38$$ ✔

For $$n = 58$$: $$r = 23$$, and $$0 \leq 23 \leq 58$$ ✔

The smallest value of $$n$$ is $$38$$.

The correct answer is Option B: 38.

We begin with the binomial expression $$\left(7^{1/5}-3^{1/10}\right)^{60}.$$ According to the binomial theorem, which states that for any real numbers $$a$$ and $$b$$ and a non-negative integer $$n$$ we have $$\left(a+b\right)^n=\sum_{k=0}^{n}\binom{n}{k}a^{\,n-k}b^{\,k},$$ the general term of the expansion of $$\left(7^{1/5}-3^{1/10}\right)^{60}$$ is obtained by setting $$a=7^{1/5},\;b=-3^{1/10},\;n=60.$$

So the $$(k+1)\text{-th}$$ term (with $$k$$ starting from $$0$$) is

$$T_{k+1}=\binom{60}{k}\left(7^{1/5}\right)^{60-k}\left(-3^{1/10}\right)^{k}.$$

We separate the sign factor $${(-1)}^{k}$$ and write the powers of $$7$$ and $$3$$ explicitly:

$$T_{k+1}=(-1)^{k}\binom{60}{k}\;7^{\,(60-k)/5}\;3^{\,k/10}.$$

For the term $$T_{k+1}$$ to be rational, both exponents $$\dfrac{60-k}{5}$$ and $$\dfrac{k}{10}$$ must be whole numbers, because $$7^{\text{integer}}\times3^{\text{integer}}$$ is an integer (apart from the possible sign) and therefore rational.

First we impose the condition

$$\frac{k}{10}\in\mathbb{Z}\quad\Longrightarrow\quad k\text{ must be divisible by }10.$$

Within the range $$0\le k\le60,$$ the values of $$k$$ satisfying this requirement are $$k=0,\,10,\,20,\,30,\,40,\,50,\,60.$$

Next we check the second condition

$$\frac{60-k}{5}\in\mathbb{Z}\quad\Longrightarrow\quad60-k\text{ must be divisible by }5.$$

If $$k$$ is already a multiple of $$10,$$ then $$60-k$$ is automatically a multiple of $$5$$ (because subtracting a multiple of $$10$$ from $$60$$ leaves a number ending in $$0$$, which is divisible by $$5$$). Thus every one of the above $$k$$-values satisfies the second condition as well.

Therefore, exactly the following $$7$$ terms are rational:

$$k=0,\;10,\;20,\;30,\;40,\;50,\;60.$$

The total number of terms in a binomial expansion with exponent $$60$$ is $$60+1=61.$$

Hence the number of irrational terms is obtained by subtracting the $$7$$ rational terms from the total number of terms:

$$\text{Number of irrational terms}=61-7=54.$$

Hence, the correct answer is Option C.

We have to find the coefficient of $$x^{10}$$ in the product

$$\left(1+x\right)^2\left(1+x^2\right)^3\left(1+x^3\right)^4.$$

First, we expand each factor completely because that will let us see every possible power along with its coefficient.

For the first factor we use the Binomial Theorem, $$\left(1+x\right)^2=\sum_{r=0}^{2}\binom{2}{r}x^{\,r},$$ so

$$\left(1+x\right)^2 = 1 + 2x + x^2.$$

For the second factor, again using the Binomial Theorem but remembering that the term in the bracket is $$x^2,$$ we get

$$\left(1+x^2\right)^3=\sum_{s=0}^{3}\binom{3}{s}x^{\,2s} = 1 + 3x^2 + 3x^4 + x^6.$$

For the third factor we do the same, treating the basic term as $$x^3:$$

$$\left(1+x^3\right)^4=\sum_{t=0}^{4}\binom{4}{t}x^{\,3t} = 1 + 4x^3 + 6x^6 + 4x^9 + x^{12}.$$

Now we must pick one term from each of the three expanded brackets such that the total power of $$x$$ is $$10.$$ Let the powers chosen from the three factors be:

$$e_1\;(from\;(1+x)^2),\quad e_2\;(from\;(1+x^2)^3),\quad e_3\;(from\;(1+x^3)^4).$$

Because the powers in the second factor come in multiples of 2 and in the third factor in multiples of 3, the only values those exponents can assume (visible in the expansions above) are

$$e_1\in\{0,1,2\},\quad e_2\in\{0,2,4,6\},\quad e_3\in\{0,3,6,9,12\}.$$

We require

$$e_1 + e_2 + e_3 = 10.$$

We check each possible choice of $$e_3$$ one by one and see whether suitable $$e_1,e_2$$ exist.

1. If $$e_3 = 0,$$ then $$e_1+e_2=10,$$ but the largest $$e_2$$ available is $$6,$$ so this case is impossible.

2. If $$e_3 = 3,$$ then $$e_1+e_2=7.$$ Because $$e_2$$ must be even, the only even value near 7 is $$6;$$ that forces $$e_1=1.$$ So one valid triple is $$(e_1,e_2,e_3)=(1,6,3).$$

3. If $$e_3 = 6,$$ then $$e_1+e_2=4.$$ The even options for $$e_2$$ are $$0,2,4.$$   • Taking $$e_2=4$$ gives $$e_1=0.$$br />   • Taking $$e_2=2$$ gives $$e_1=2.$$br /> $$e_2=0$$ would force $$e_1=4,$$ which is not allowed, so we get two more triples: $$(e_1,e_2,e_3)=(0,4,6)\quad\text{and}\quad(2,2,6).$$

4. If $$e_3 = 9,$$ then $$e_1+e_2=1.$$ Since $$e_2$$ is even, the only choice is $$e_2=0,$$ giving $$e_1=1.$$ Thus we obtain $$(e_1,e_2,e_3)=(1,0,9).$$

5. If $$e_3 = 12,$$ then $$e_1+e_2=-2,$$ impossible.

Hence the only admissible exponent triples are

$$ \begin{aligned} (1,6,3),\quad (0,4,6),\quad (2,2,6),\quad (1,0,9). \end{aligned} $$

For each triple we multiply the corresponding coefficients taken from the three expansions.

For $$(e_1,e_2,e_3)=(1,6,3):$$ coefficient $$= \color{blue}{2}\times\color{blue}{1}\times\color{blue}{4}=8.$$ (The numbers in blue are from $$2x,\;x^6,\;4x^3.$)

For $$(e_1,e_2,e_3)=(0,4,6):$$ coefficient $$= 1$$\times$$3$$\times$$6 = 18.$$

For $$(e_1,e_2,e_3)=(2,2,6):$$ coefficient $$= 1$$\times$$3$$\times$$6 = 18.$$

For $$(e_1,e_2,e_3)=(1,0,9):$$ coefficient $$= 2$$\times$$1$$\times$$4 = 8.$$

Finally, we add all these contributions because each gives an independent term with power $$10$$:

$$8 + 18 + 18 + 8 = 52.$$

Hence, the coefficient of $$x^{10}$$ in the given expansion is $$52.$$ Therefore, the correct answer is Option A.

We have to find the sum of the coefficients of all those terms whose powers of $$x$$ are odd in the expression

Put $$y=\sqrt{x^{3}-1}$$ just for convenience. The given expression becomes

First recall the binomial formula:

Using it for $$(x+y)^{5}$$ we get

For $$(x-y)^{5}$$ every odd power of $$y$$ picks up a minus sign, so we obtain

Now add the two expansions term-by-term:

All parts containing odd powers of $$y$$ cancel, while the even ones double. Hence

Next rewrite everything solely in powers of $$x$$. Because $$y=\sqrt{x^{3}-1}$$, we have $$y^{2}=x^{3}-1$$ and therefore $$y^{4}=(y^{2})^{2}=(x^{3}-1)^{2}=x^{6}-2x^{3}+1.$$ Substitute these:

Work out each product fully:

Collect every term:

The odd powers of $$x$$ present here are $$x^{7},x^{5},x^{3}$$ and $$x^{1}$$. Their respective coefficients are

Add these coefficients:

So the sum of the coefficients of all odd-degree terms is $$2$$.

Hence, the correct answer is Option A.

We have to find the coefficient of $$x^{2}$$ in

$$\left(2 - x^{2}\right)\,\Big\{(1 + 2x + 3x^{2})^{6} + (1 - 4x^{2})^{6}\Big\}.$$

For convenience let us denote

$$A(x)= (1 + 2x + 3x^{2})^{6}, \qquad B(x)= (1 - 4x^{2})^{6}.$$

Then the required product is$$(2 - x^{2})\,[A(x)+B(x)].$$

When two polynomials are multiplied, the coefficient of a specific power (here $$x^{2}$$) can be obtained by pairing terms whose powers add up to that exponent. Multiplying $$2 - x^{2}$$ with $$A(x)+B(x)$$ gives

$$\bigl(2 - x^{2}\bigr)\bigl[A(x)+B(x)\bigr] = 2\,\bigl[A(x)+B(x)\bigr] \;-\; x^{2}\,\bigl[A(x)+B(x)\bigr].$$

Therefore the coefficient of $$x^{2}$$ in the final expression is

$$2 \times \bigl[\text{coefficient of }x^{2}\text{ in }A(x)+B(x)\bigr]\;-\; 1 \times \bigl[\text{coefficient of }x^{0}\text{ in }A(x)+B(x)\bigr].$$

So we need two numbers:

1. $$C_{0} =$$ coefficient of $$x^{0}$$ (constant term) in $$A(x)+B(x).$$
2. $$C_{2} =$$ coefficient of $$x^{2}$$ in $$A(x)+B(x).$$

We shall evaluate them one by one.

Constant terms

In any expansion $$(1 + \alpha)^{n},$$ the constant term is simply $$1^{n}=1.$$

Hence
$$\text{constant term of }A(x)=1,$$ because $$A(x)=(1+2x+3x^{2})^{6}$$ gives $$1$$ when we choose the $$1$$ from each of the six factors.

Similarly,
$$\text{constant term of }B(x)=1,$$ because $$B(x)=(1-4x^{2})^{6}$$ also gives $$1$$ when we choose the $$1$$ from every factor.

Thus

$$C_{0}=1+1=2.$$

Coefficient of $$x^{2}$$ in $$A(x)=(1+2x+3x^{2})^{6}$$

To obtain $$x^{2}$$ from the multinomial expansion we can proceed term-by-term:

• Choose the linear term $$2x$$ from exactly two of the six brackets and the constant $$1$$ from the remaining four. Each such choice contributes $$\bigl(2x\bigr)^{2}(1)^{4}=4x^{2}.$$ The number of ways is $$\binom{6}{2}=15.$$ Hence the contribution is $$15\times4=60.$$
• Choose the quadratic term $$3x^{2}$$ from exactly one of the six brackets and the constant $$1$$ from the other five. Each such choice contributes $$3x^{2}.$$ The number of ways is $$\binom{6}{1}=6.$$ Hence the contribution is $$6\times3=18.$$
Adding these two contributions we get

$$\text{coefficient of }x^{2}\text{ in }A(x)=60+18=78.$$

Coefficient of $$x^{2}$$ in $$B(x)=(1-4x^{2})^{6}$$

Here every non-constant term already has an even power of $$x$$ because the middle term is $$-4x^{2}.$$ To get exactly $$x^{2}$$ we must pick the term $$-4x^{2}$$ from one factor and the constant $$1$$ from the remaining five factors. Thus

$$\text{coefficient of }x^{2}\text{ in }B(x) =\binom{6}{1}\times(-4)=6\times(-4)=-24.$$

Adding the coefficients from $$A(x)$$ and $$B(x)$$ we obtain

$$C_{2}=78+(-24)=54.$$

Putting the pieces together

Recall that the coefficient required is

$$2C_{2}-C_{0}=2\times54-2=108-2=106.$$

Hence, the correct answer is Option D.

$$S = \underbrace{\sum_{k=1}^{10} {}^{21}C_{k}}_{S_1} - \underbrace{\sum_{k=1}^{10} {}^{10}C_{k}}_{S_2}$$

$$\sum_{k=0}^{10} {}^{10}C_{k} = 2^{10}$$

$$S_2 = \sum_{k=1}^{10} {}^{10}C_{k} = 2^{10} - {}^{10}C_{0} = \mathbf{2^{10} - 1}$$

For $$n = 21$$, the total sum is $$\sum_{k=0}^{21} {}^{21}C_{k} = 2^{21}$$

Since $${}^{n}C_{k} = {}^{n}C_{n-k}$$, the sum of the first half of the terms is equal to the second half:

$$\sum_{k=0}^{10} {}^{21}C_{k} = \sum_{k=11}^{21} {}^{21}C_{k} = \frac{2^{21}}{2} = 2^{20}$$

$$S_1 = \sum_{k=1}^{10} {}^{21}C_{k} = 2^{20} - {}^{21}C_{0} = \mathbf{2^{20} - 1}$$

$$S = (2^{20} - 1) - (2^{10} - 1)$$

$$S = 2^{20} - 2^{10}$$

We have to find the remainder obtained when the very large power $$27^{999}$$ is divided by $$7$$. In other words, we must evaluate $$27^{999}\pmod 7.$$ Working entirely in modular arithmetic keeps the numbers small and the calculations manageable.

First, reduce the base modulo $$7$$:

$$27 = 7\times 3 + 6 \; \Longrightarrow \; 27 \equiv 6 \pmod 7.$$

So the original expression can be rewritten as

$$27^{999} \equiv 6^{999}\pmod 7.$$

Now, instead of tackling the exponent $$999$$ directly, let us examine the powers of $$6$$ modulo $$7$$ to see whether they follow a simple repeating pattern.

Compute the first few powers:

$$6^1 \equiv 6 \pmod 7,$$

$$6^2 = 36 \equiv 36 - 35 = 1 \pmod 7,$$

$$6^3 = 6^2\cdot 6 \equiv 1\cdot 6 = 6 \pmod 7,$$

$$6^4 = (6^2)^2 \equiv 1^2 = 1 \pmod 7.$$

From these results we observe the clear two-term cycle

$$6,\,1,\,6,\,1,\,\dots$$

In particular,

$$\bigl(6^1\bigr)\equiv 6,\qquad \bigl(6^2\bigr)\equiv 1,$$

and therefore

$$6^{\text{odd exponent}}\equiv 6,\qquad 6^{\text{even exponent}}\equiv 1 \pmod 7.$$

The exponent we need is $$999$$, which is odd. Hence

$$6^{999}\equiv 6 \pmod 7.$$

But $$6$$ is already less than $$7,$$ so it is itself the remainder.

Therefore, when $$27^{999}$$ is divided by $$7,$$ the remainder is $$6$$.

Hence, the correct answer is Option C.

The task is to locate the coefficient of the term whose power of $$x$$ is $$-5$$ in the expansion of

$$\left(\dfrac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\dfrac{x-1}{x-x^{\frac12}}\right)^{10},\qquad x\neq0,\;x\neq1.$$

First the two fractions inside will be simplified separately.

For the first fraction let $$y=x^{\frac13}.$$ Then

$$x^{\frac23}=y^{2},\qquad x+1=y^{3}+1.$$

The fraction becomes

$$\dfrac{y^{3}+1}{\,y^{2}-y+1\,}.$$

Using the factorisation $$y^{3}+1=(y+1)(y^{2}-y+1)$$ gives

$$\dfrac{(y+1)(y^{2}-y+1)}{\,y^{2}-y+1\,}=y+1.$$

Undoing the substitution, the first fraction is

$$x^{\frac13}+1.$$

For the second fraction put $$z=x^{\frac12}.$$ Then $$x=z^{2},$$ so

$$x-1=z^{2}-1=(z-1)(z+1),$$ $$x-x^{\frac12}=z^{2}-z=z(z-1).$$

Hence

$$\dfrac{z^{2}-1}{\,z^{2}-z\,}= \dfrac{(z-1)(z+1)}{\,z(z-1)\,}= \dfrac{z+1}{z}=1+\dfrac1z.$$

Returning to $$x,$$ the second fraction is

$$1+x^{-\frac12}.$$

The expression inside the large parentheses therefore simplifies to

$$(x^{\frac13}+1)-(1+x^{-\frac12})=x^{\frac13}-x^{-\frac12}.$$

The whole expression becomes

$$(x^{\frac13}-x^{-\frac12})^{10}.$$

Denote $$a=x^{\frac13},\;b=-x^{-\frac12};$$ then the binomial theorem gives the general term

$$T_{k}= \binom{10}{k}a^{\,10-k}b^{\,k}= \binom{10}{k}(x^{\frac13})^{10-k}\bigl(-x^{-\frac12}\bigr)^{k},\qquad k=0,1,\dots,10.$$

The power of $$x$$ in this term is

$$\left(10-k\right)\left(\dfrac13\right)+k\!\left(-\dfrac12\right)=\dfrac{10-k}{3}-\dfrac{k}{2}.$$

Putting everything over the common denominator 6,

$$\dfrac{2(10-k)-3k}{6}= \dfrac{20-2k-3k}{6}= \dfrac{20-5k}{6}.$$

We want this exponent to equal $$-5,$$ so set

$$\dfrac{20-5k}{6}=-5.$$

Multiplying by 6:

$$20-5k=-30.$$

Subtract 20:

$$-5k=-50.$$

Divide by $$-5$$:

$$k=10.$$

Only the term with $$k=10$$ contributes to the coefficient of $$x^{-5}.$$ For this value,

$$T_{10}= \binom{10}{10}(x^{\frac13})^{0}\bigl(-x^{-\frac12}\bigr)^{10} =1\cdot1\cdot\bigl[(-1)^{10}x^{-5}\bigr]=x^{-5}.$$

The numerical coefficient is therefore $$1.$$

Hence, the correct answer is Option C.

We have to evaluate the finite sum

$$\sum_{r=1}^{15} r^{2}\left(\frac{{}^{15}C_{r}}{{}^{15}C_{r-1}}\right).$$

First we simplify the ratio of the two consecutive binomial coefficients. By definition

$${}^{n}C_{k}= \frac{n!}{k!\,(n-k)!}.$$

So, taking $$n=15$$, we can write

$${}^{15}C_{r}= \frac{15!}{r!\,(15-r)!}, \qquad {}^{15}C_{\,r-1}= \frac{15!}{(r-1)!\,\bigl(15-(r-1)\bigr)!} = \frac{15!}{(r-1)!\,(16-r)!}.$$

Now form the required ratio:

$$\frac{{}^{15}C_{r}}{{}^{15}C_{r-1}} =\frac{\dfrac{15!}{r!\,(15-r)!}} {\dfrac{15!}{(r-1)!\,(16-r)!}} =\frac{(r-1)!\,(16-r)!}{r!\,(15-r)!}.$$

Break this fraction into two separate factors:

$$\frac{(r-1)!}{r!}\times\frac{(16-r)!}{(15-r)!}.$$

We know $$r! = r\,(r-1)!$$, hence

$$\frac{(r-1)!}{r!}=\frac{1}{r}.$$

Similarly, $$(16-r)! = (16-r)\,(15-r)!$$, so

$$\frac{(16-r)!}{(15-r)!}=16-r.$$

Multiplying these two results we obtain

$$\frac{{}^{15}C_{r}}{{}^{15}C_{r-1}}=\frac{1}{r}\,(16-r)=\frac{16-r}{r}.$$

Put this back into the original sum. Each term inside the summation becomes

$$r^{2}\left(\frac{16-r}{r}\right)=r^{2}\cdot\frac{16-r}{r}=r\,(16-r).$$

Thus the given sum simplifies to

$$\sum_{r=1}^{15} r\,(16-r).$$

Distribute the product inside the summation:

$$\sum_{r=1}^{15} r\,(16-r)=\sum_{r=1}^{15}\bigl(16r-r^{2}\bigr) =16\sum_{r=1}^{15}r-\sum_{r=1}^{15}r^{2}.$$

Now we evaluate the two standard series that appear.

Formula for the sum of the first $$n$$ natural numbers:

$$\sum_{r=1}^{n} r =\frac{n(n+1)}{2}.$$

Taking $$n=15$$ gives

$$\sum_{r=1}^{15} r=\frac{15\times16}{2}=120.$$

Formula for the sum of the squares of the first $$n$$ natural numbers:

$$\sum_{r=1}^{n} r^{2}= \frac{n(n+1)(2n+1)}{6}.$$

With $$n=15$$ we get

$$\sum_{r=1}^{15} r^{2}= \frac{15\times16\times31}{6}.$$

Compute this step by step:

$$15\times16=240,$$

$$240\times31 = 7440,$$

$$\frac{7440}{6}=1240.$$

Substitute these two results back into our expression:

$$16\sum_{r=1}^{15} r-\sum_{r=1}^{15} r^{2} =16\times120-1240 =1920-1240 =680.$$

So, the entire sum equals

$$680.$$

Hence, the correct answer is Option D.

We start with the given finite sum

defined for all real $$x$$ with $$x\neq -1$$. It is convenient to write the same expression with an index so that every term follows the same pattern:

At this point we recall the algebraic identity for the difference of two powers:

This identity is valid for all real or complex numbers $$a,b$$ and every non-negative integer $$n$$. We now apply it with the specific choices

Substituting these values in the general formula, we obtain

Inside the large parentheses we plainly have

Hence the right-hand side simplifies to

The last sum is exactly our original $$S$$, therefore we have established the compact form

Our task is to find $$a_{17}$$, the coefficient of $$x^{17}$$ when $$S$$ is expanded as a polynomial in $$x$$. The representation we have just derived makes this easy:

• The term $$(1+x)^{2017}$$ contains a contribution $$\binom{2017}{17}x^{17}$$.
• The term $$-\,x^{2017}$$ does not contain any $$x^{17}$$ part, because its only term is $$-x^{2017}$$.

Therefore the entire coefficient of $$x^{17}$$ in $$S$$ is simply

Thus

Comparing with the options, we see that this value corresponds to Option A.

Hence, the correct answer is Option A.

We begin with the binomial expression $$\left(x^{1/3} + \frac{1}{2x^{1/3}}\right)^{18}.$$ To expand it we use the binomial theorem, which states that for any real numbers $$u$$ and $$v$$ and a non-negative integer $$n$$, we have $$ (u+v)^n=\sum_{r=0}^{n}\binom{n}{r}u^{\,n-r}v^{\,r}. $$

Here $$u= x^{1/3},\qquad v=\frac{1}{2x^{1/3}},\qquad n=18.$$ So the general term of the expansion is

$$ T_{r+1}= \binom{18}{r}\left(x^{1/3}\right)^{18-r}\left(\frac{1}{2x^{1/3}}\right)^{r}, \qquad 0\le r\le 18. $$

First we simplify the power of $$x$$ inside this term. We have

$$ \left(x^{1/3}\right)^{18-r}=x^{(18-r)/3},\qquad \left(\frac{1}{2x^{1/3}}\right)^{r}= \frac{1}{2^{r}}\,x^{-r/3}. $$

Multiplying these two powers of $$x$$, we obtain

$$ x^{(18-r)/3}\,x^{-r/3}=x^{\frac{18-r-r}{3}}=x^{\frac{18-2r}{3}}. $$

Thus the complete general term is

$$ T_{r+1}= \binom{18}{r}\frac{1}{2^{r}}\,x^{\frac{18-2r}{3}}. $$

We need the coefficients of the terms $$x^{-2}$$ and $$x^{-4}$$. So we equate the exponent $$\dfrac{18-2r}{3}$$ with $$-2$$ and $$-4$$ successively.

For $$x^{-2}$$:

$$ \frac{18-2r}{3}=-2 \;\;\Longrightarrow\;\; 18-2r=-6 \;\;\Longrightarrow\;\; 2r=24 \;\;\Longrightarrow\;\; r=12. $$

For $$x^{-4}$$:

$$ \frac{18-2r}{3}=-4 \;\;\Longrightarrow\;\; 18-2r=-12 \;\;\Longrightarrow\;\; 2r=30 \;\;\Longrightarrow\;\; r=15. $$

Hence

• The coefficient $$m$$ of $$x^{-2}$$ is obtained from $$r=12$$: $$ m=\binom{18}{12}\frac{1}{2^{12}}. $$
• The coefficient $$n$$ of $$x^{-4}$$ is obtained from $$r=15$$: $$ n=\binom{18}{15}\frac{1}{2^{15}}. $$

We now evaluate the required binomial coefficients.

Because $$\binom{18}{12}=\binom{18}{6}$$ (symmetry property), we calculate

$$ \binom{18}{6}= \frac{18!}{6!\,12!} =\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13}{6\cdot5\cdot4\cdot3\cdot2\cdot1} =18\,564. $$

Similarly

$$ \binom{18}{15}=\binom{18}{3} =\frac{18\cdot17\cdot16}{3\cdot2\cdot1} =816. $$

Therefore

$$ m=18\,564\left(\frac{1}{2}\right)^{12},\qquad n=816\left(\frac{1}{2}\right)^{15}. $$

We now find their ratio:

$$$ \frac{m}{n}= \frac{18\,564\left(\dfrac{1}{2}\right)^{12}} {816\left(\dfrac{1}{2}\right)^{15}} \;=\; \frac{18\,564}{816}\times\left(\frac{1}{2}\right)^{12-15} \;=\; \frac{18\,564}{816}\times 2^{3}. $$$

Calculating step by step,

$$ \frac{18\,564}{816}=22.75=\frac{1\,547}{68}, $$

$$ \left(\frac{18\,564}{816}\right)\times 2^{3} =\frac{1\,547}{68}\times 8 =\frac{12\,376}{68} =182. $$

So we obtain

$$ \frac{m}{n}=182. $$

Hence, the correct answer is Option B.

We begin with the multinomial expression $$\left(1-\dfrac{2}{x}+\dfrac{4}{y^2}\right)^n.$$

In every term of its expansion the exponents of the three individual parts $$1,\;-\dfrac{2}{x},\;\dfrac{4}{y^2}$$ must add up to the power $$n.$$

Let those exponents be $$r,\;s,\;t$$ respectively. We therefore have the condition $$r+s+t=n,$$ where $$r,\;s,\;t$$ are non-negative integers.

The total number of ordered triples $$(r,s,t)$$ satisfying $$r+s+t=n$$ equals the number of non-negative integral solutions of that equation. A standard result from combinatorics states that the number of such solutions is

$$\binom{n+3-1}{3-1}=\binom{n+2}{2}.$$

It is given that the expansion contains exactly 28 distinct terms, so

$$\binom{n+2}{2}=28.$$

Using the definition $$\displaystyle \binom{n+2}{2}=\dfrac{(n+2)(n+1)}{2},$$ we write

$$\dfrac{(n+2)(n+1)}{2}=28.$$

Multiplying both sides by 2, we obtain

$$(n+2)(n+1)=56.$$

Now we expand the product on the left:

$$n^2+3n+2=56.$$

Subtracting 56 from both sides gives

$$n^2+3n+2-56=0$$ $$\Rightarrow n^2+3n-54=0.$$

We look for factors of $$-54$$ whose sum is $$3.$$ The factor pair $$9$$ and $$-6$$ works, so

$$(n+9)(n-6)=0.$$

This yields two possible values $$n=-9$$ or $$n=6.$$ Since $$n$$ must be a non-negative integer, we take

$$n=6.$$

Now we need the sum of the coefficients of all terms in the expansion. A well-known fact is: “The sum of the coefficients in a multinomial expansion is obtained by substituting every variable by $$1.$$”

Here each term in $$\left(1-\dfrac{2}{x}+\dfrac{4}{y^2}\right)^6$$ contains the factors $$1,\;x^{-1},\;y^{-2}.$$ If we put $$x=1$$ and $$y=1,$$ every power of $$x$$ and $$y$$ becomes $$1,$$ leaving behind only the numerical coefficients. Thus

$$\text{Sum of coefficients}= \left(1-\dfrac{2}{1}+\dfrac{4}{1^2}\right)^6 = (1-2+4)^6 = (3)^6 = 729.$$

Hence, the correct answer is Option B.

We consider the binomial expansion of $$\bigl(1+x\bigr)^n.$$

The general term (counting from 0) is written first:

$$T_{r+1}= \binom{n}{r}\,x^{\,r},$$

where $$r=0,1,2,\dots,n.$$ The numerical coefficient of this term is simply $$\binom{n}{r}.$$

According to the question, the numerical coefficients of three successive terms are in the ratio $$1:7:42.$$ Let these three successive terms correspond to the indices $$r,\;r+1,\;r+2.$$ Then we have

$$\binom{n}{r} : \binom{n}{r+1} : \binom{n}{r+2}=1:7:42.$$

To translate the verbal ratio into equations, we equate the successive ratios of these coefficients:

First ratio:

$$\frac{\binom{n}{r+1}}{\binom{n}{r}}=7.$$

Second ratio:

$$\frac{\binom{n}{r+2}}{\binom{n}{r+1}}=\frac{42}{7}=6.$$

Now we recall the standard identity for consecutive binomial coefficients:

$$\frac{\binom{n}{k+1}}{\binom{n}{k}}=\frac{n-k}{k+1}.$$

Applying this identity to each of the above ratios, we obtain two algebraic equations.

For the first ratio:

$$\frac{\binom{n}{r+1}}{\binom{n}{r}}=\frac{n-r}{r+1}=7.$$

Hence

$$n-r=7(r+1).$$

Simplifying,

$$n-r=7r+7 \;\;\Longrightarrow\;\; n=8r+7.$$

For the second ratio we write

$$\frac{\binom{n}{r+2}}{\binom{n}{r+1}}=\frac{n-(r+1)}{r+2}=6.$$

Simplifying the numerator gives

$$\frac{n-r-1}{r+2}=6.$$

Cross-multiplying, we find

$$n-r-1=6(r+2).$$

Expanding the right side,

$$n-r-1=6r+12.$$

Hence

$$n=7r+13.$$

We now have two expressions for $$n:$$

$$n=8r+7 \quad\text{and}\quad n=7r+13.$$

Equating these,

$$8r+7=7r+13.$$

Subtracting $$7r$$ from both sides,

$$r+7=13.$$

So

$$r=6.$$

Substituting $$r=6$$ back into either expression for $$n,$$ say $$n=8r+7,$$ we get

$$n=8(6)+7=48+7=55.$$

Thus the three successive terms are the ones with indices

$$r=6,\; r+1=7,\; r+2=8.$$

The first of these, $$r=6,$$ corresponds to the term

$$T_{r+1}=T_{6+1}=T_7.$$

Since term counting in binomial expansions starts with $$T_1$$ (the $$r=0$$ term), $$T_7$$ is called the 7th term of the expansion.

Hence, the correct answer is Option D.

We have to add up the coefficients of only those terms in the expansion of $$(1-2\sqrt{x})^{50}$$ that contain an integral (whole-number) power of $$x$$.

First recall the binomial theorem, which states that for any real numbers $$a$$ and $$b$$ and any non-negative integer $$n$$, we have

$$ (a+b)^n \;=\; \sum_{k=0}^{n} \binom{n}{k}\,a^{\,n-k}\,b^{\,k}. $$

To expand $$(1-2\sqrt{x})^{50}$$ we put $$a=1$$ and $$b=-2\sqrt{x}$$, giving the general (the $$k^{\text{th}}$$) term

$$ T_k \;=\; \binom{50}{k}\,(1)^{\,50-k}\,(-2\sqrt{x})^{k} \;=\; \binom{50}{k}\,(-1)^{k}\,2^{k}\,x^{k/2}. $$

The power of $$x$$ that appears in this term is $$k/2$$. For this power to be an integer, $$k/2$$ must itself be an integer, which is possible only when $$k$$ is even. So we set $$k=2m$$, where $$m$$ runs over all integers from $$0$$ to $$25$$:

$$ k = 2m,\qquad m = 0,1,2,\dots ,25. $$

Substituting $$k=2m$$ into the expression for $$T_k$$ gives

$$ T_{2m} \;=\; \binom{50}{2m}\, (-1)^{2m}\,2^{2m}\,x^{m}. $$

Because $$(-1)^{2m}=1$$ for every integer $$m$$, the coefficient of $$x^{m}$$ in this term is simply

$$ \binom{50}{2m}\,2^{2m}. $$

The required sum (call it $$S$$) is the sum of these coefficients for all admissible values of $$m$$:

$$ S \;=\; \sum_{m=0}^{25} \binom{50}{2m}\,2^{2m}. $$

To evaluate this quickly, we use a standard identity for separating even and odd terms of a binomial expansion.

First write the expansion of $$(1+2)^{50}$$ and $$(1-2)^{50}$$ using the binomial theorem:

$$ (1+2)^{50} \;=\; \sum_{k=0}^{50} \binom{50}{k}\,1^{\,50-k}\,2^{k} \;=\; \sum_{k=0}^{50} \binom{50}{k}\,2^{k}, $$

$$ (1-2)^{50} \;=\; \sum_{k=0}^{50} \binom{50}{k}\,1^{\,50-k}\,(-2)^{k} \;=\; \sum_{k=0}^{50} (-1)^{k}\,\binom{50}{k}\,2^{k}. $$

Now add these two expansions term by term:

$$ (1+2)^{50} + (1-2)^{50} \;=\; \sum_{k=0}^{50} \binom{50}{k}\,2^{k} + \sum_{k=0}^{50} (-1)^{k}\,\binom{50}{k}\,2^{k}. $$

For every odd $$k$$, the two corresponding terms are equal in magnitude but opposite in sign and therefore cancel. For every even $$k$$, the signs are the same and the terms double. Hence the sum simplifies to

$$ (1+2)^{50} + (1-2)^{50} \;=\; 2 \sum_{k\;\text{even}} \binom{50}{k}\,2^{k} \;=\; 2 \sum_{m=0}^{25} \binom{50}{2m}\,2^{2m}. $$

Comparing with our definition of $$S$$, we see that

$$ 2S \;=\; (1+2)^{50} + (1-2)^{50}. $$

But $$1+2 = 3$$ and $$1-2 = -1$$, so

$$ (1+2)^{50} = 3^{50},\qquad (1-2)^{50} = (-1)^{50} = 1. $$

Therefore

$$ 2S \;=\; 3^{50} + 1, $$

and dividing by $$2$$ gives

$$ S \;=\; \frac{3^{50} + 1}{2}. $$

This expression is exactly option B.

Hence, the correct answer is Option B.

We need to find the term independent of $$ x $$ in the expansion of $$ \left(1 - \frac{1}{x} + 3x^5\right)\left(2x^2 - \frac{1}{x}\right)^8 $$. This term will have an exponent of zero for $$ x $$. We start by expanding the second part using the binomial theorem.

The binomial expansion of $$ \left(2x^2 - \frac{1}{x}\right)^8 $$ is given by:

$$ \left(2x^2 - \frac{1}{x}\right)^8 = \sum_{k=0}^{8} \binom{8}{k} (2x^2)^{8-k} \left(-\frac{1}{x}\right)^k $$

Simplify each term:

$$ \binom{8}{k} (2x^2)^{8-k} \left(-\frac{1}{x}\right)^k = \binom{8}{k} 2^{8-k} x^{2(8-k)} \cdot (-1)^k x^{-k} = \binom{8}{k} 2^{8-k} (-1)^k x^{16 - 2k - k} = \binom{8}{k} 2^{8-k} (-1)^k x^{16 - 3k} $$

So, the general term is $$ T_k = \binom{8}{k} 2^{8-k} (-1)^k x^{16 - 3k} $$.

Now, multiply this by the first part $$ \left(1 - \frac{1}{x} + 3x^5\right) $$:

$$ \left(1 - \frac{1}{x} + 3x^5\right) \sum_{k=0}^{8} T_k = \left(1 - \frac{1}{x} + 3x^5\right) \sum_{k=0}^{8} \binom{8}{k} 2^{8-k} (-1)^k x^{16 - 3k} $$

To find the constant term, we look for combinations where the total exponent of $$ x $$ is zero. This happens when:

1. Multiplying by 1 (exponent 0) and the term from the expansion has exponent $$ 16 - 3k = 0 $$.
2. Multiplying by $$ -\frac{1}{x} $$ (exponent -1) and the term from the expansion has exponent such that $$ (16 - 3k) + (-1) = 0 $$.
3. Multiplying by $$ 3x^5 $$ (exponent 5) and the term from the expansion has exponent such that $$ (16 - 3k) + 5 = 0 $$.

Solve each equation for integer $$ k $$ (since $$ k $$ must be an integer between 0 and 8):

1. $$ 16 - 3k = 0 $$ → $$ 3k = 16 $$ → $$ k = \frac{16}{3} \approx 5.333 $$, not an integer. No solution.
2. $$ 16 - 3k - 1 = 0 $$ → $$ 15 - 3k = 0 $$ → $$ 3k = 15 $$ → $$ k = 5 $$, which is an integer and within range.
3. $$ 16 - 3k + 5 = 0 $$ → $$ 21 - 3k = 0 $$ → $$ 3k = 21 $$ → $$ k = 7 $$, which is an integer and within range.

So, only cases 2 and 3 give valid $$ k $$ values. Now compute their contributions.

Case 2: $$ k = 5 $$, multiplied by $$ -\frac{1}{x} $$

First, find $$ T_5 $$:

$$ T_5 = \binom{8}{5} 2^{8-5} (-1)^5 x^{16 - 3 \times 5} = \binom{8}{5} 2^{3} (-1)^5 x^{16-15} = \binom{8}{5} \cdot 8 \cdot (-1) x^{1} $$

Calculate $$ \binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 $$, so:

$$ T_5 = 56 \cdot 8 \cdot (-1) \cdot x = -448x $$

Now multiply by $$ -\frac{1}{x} $$:

$$ \left(-\frac{1}{x}\right) \times (-448x) = (-1) \times (-448) \times \frac{x}{x} = 448 \times 1 = 448 $$

Contribution is 448.

Case 3: $$ k = 7 $$, multiplied by $$ 3x^5 $$

First, find $$ T_7 $$:

$$ T_7 = \binom{8}{7} 2^{8-7} (-1)^7 x^{16 - 3 \times 7} = \binom{8}{7} 2^{1} (-1)^7 x^{16-21} = \binom{8}{7} \cdot 2 \cdot (-1) x^{-5} $$

Calculate $$ \binom{8}{7} = 8 $$, so:

$$ T_7 = 8 \cdot 2 \cdot (-1) \cdot x^{-5} = -16x^{-5} $$

Now multiply by $$ 3x^5 $$:

$$ (3x^5) \times (-16x^{-5}) = 3 \times (-16) \times x^{5 + (-5)} = -48 \times x^0 = -48 $$

Contribution is -48.

Sum the contributions: $$ 448 + (-48) = 400 $$.

Hence, the term independent of $$ x $$ is 400.

Comparing with the options:

A. $$-496$$

B. $$-400$$

C. $$496$$

D. $$400$$

So, the correct answer is Option D.

Let $$y = 1+x$$, which implies $$x = y-1$$.

The equation becomes $$1 + (y-1)^4 + (y-1)^5 = \sum_{i=0}^{5} a_i y^i$$

Since the RHS is a polynomial in $$y$$, $$a_2$$ is specifically the coefficient of the $$y^2$$ term on the LHS.

The $$y^2$$ term in $$(y-1)^4$$ is $$\binom{4}{2}y^2(-1)^2 = 6y^2$$. The $$y^2$$ term in $$(y-1)^5$$ is $$\binom{5}{2}y^2(-1)^3 = -10y^2$$.

$$a_2 = 6 - 10 = \mathbf{-4}$$

First we write the given expression and recall the binomial theorem. We have $$\bigl(1+ax+bx^2\bigr)\,(1-2x)^{18}.$$ According to the binomial theorem,

$$ (1-2x)^{18} \;=\; \sum_{k=0}^{18} \binom{18}{k}(1)^{18-k}(-2x)^k \;=\; \sum_{k=0}^{18} (-1)^k 2^k \binom{18}{k}\,x^k. $$

For convenience we denote $$ c_k \;=\; (-1)^k 2^k \binom{18}{k}, $$ so that the series becomes $$\displaystyle\sum_{k=0}^{18} c_k x^k.$$

Now we multiply this series by $$1+ax+bx^2$$. Every power of $$x$$ in the product arises by picking one term from $$1+ax+bx^2$$ and one term from the series, and adding all such contributions.

Coefficient of $$x^3$$.
To obtain $$x^3$$ we have three possibilities:

• The factor $$1$$ from the first bracket and the term $$c_3x^3$$ from the series.
• The factor $$ax$$ and the term $$c_2x^2$$  (because $$x\cdot x^2=x^3$$).
• The factor $$bx^2$$ and the term $$c_1x$$.

Hence the coefficient of $$x^3$$ is $$ c_3 + a\,c_2 + b\,c_1. $$ The question says this coefficient is zero, so

$$ c_3 + a\,c_2 + b\,c_1 = 0. \quad -(1) $$

Coefficient of $$x^4$$.
In the same way $$x^4$$ can arise from:

• $$1$$ with $$c_4x^4,$$
• $$ax$$ with $$c_3x^3,$$
• $$bx^2$$ with $$c_2x^2.$$

Therefore the coefficient of $$x^4$$ is $$ c_4 + a\,c_3 + b\,c_2, $$ and the condition “coefficient is zero” gives

$$ c_4 + a\,c_3 + b\,c_2 = 0. \quad -(2) $$

Next we evaluate the required $$c_k$$ one by one:

Using $$c_k = (-1)^k 2^k\binom{18}{k},$$

$$ \begin{aligned} c_1 &= (-1)^1 2^1\binom{18}{1}= -2\cdot18 = -36,\\[4pt] c_2 &= (+1) 2^2\binom{18}{2}= 4\cdot153 = 612,\\[4pt] c_3 &= (-1)^3 2^3\binom{18}{3}= -8\cdot816 = -6528,\\[4pt] c_4 &= (+1) 2^4\binom{18}{4}= 16\cdot3060 = 48960. \end{aligned} $$

Substituting these numbers into equation (1) we get

$$ -6528 + a\,(612) + b\,(-36) = 0, $$ so $$ 612a - 36b - 6528 = 0. $$ Dividing every term by $$12$$ gives

$$ 51a - 3b = 544. \quad -(3) $$

Now substituting the same numbers into equation (2) yields

$$ 48960 + a\,(-6528) + b\,(612) = 0, $$ which simplifies to

$$ -6528a + 612b + 48960 = 0. $$ Dividing every term by $$12$$ gives

$$ -544a + 51b = -4080. \quad -(4) $$

We now have the linear system

$$ \begin{cases} 51a - 3b = 544,\\ -544a + 51b = -4080. \end{cases} $$

To eliminate $$b$$, we multiply equation (3) by $$17$$ so that the coefficient of $$b$$ matches that in equation (4):

$$ 17(51a - 3b) = 17\cdot544 \;\;\Longrightarrow\;\; 867a - 51b = 9248. \quad -(5) $$

Now we add equations (4) and (5):

$$ (867a - 51b) + (-544a + 51b) = 9248 + (-4080). $$ The $$b$$ terms cancel, leaving

$$ 323a = 5168 \;\;\Longrightarrow\;\; a = \frac{5168}{323}. $$

Since $$323\times16 = 5168$$, we have

$$ a = 16. $$

Substituting $$a=16$$ back into equation (3):

$$ 51(16) - 3b = 544 \;\;\Longrightarrow\;\; 816 - 3b = 544, $$ so

$$ -3b = 544 - 816 = -272 \;\;\Longrightarrow\;\; b = \frac{272}{3}. $$

Thus the ordered pair $$(a,b)$$ equals $$ \left(16,\;\frac{272}{3}\right). $$

Hence, the correct answer is Option B.

The given expression is $$(1 + x^n + x^{253})^{10}$$, and we need to find the coefficient of $$x^{1012}$$ where $$n$$ is a positive integer with $$n \leq 22$$.

In the expansion of $$(1 + x^n + x^{253})^{10}$$, each term arises from selecting one of the three options—$$1$$, $$x^n$$, or $$x^{253}$$—from each of the 10 factors. The general term is given by the multinomial coefficient:

$$\text{term} = \binom{10}{a, b, c} \cdot (1)^a \cdot (x^n)^b \cdot (x^{253})^c = \binom{10}{a, b, c} \cdot x^{n b + 253 c}$$

where $$a$$, $$b$$, and $$c$$ are non-negative integers satisfying $$a + b + c = 10$$, and $$\binom{10}{a, b, c} = \frac{10!}{a! b! c!}$$.

We require the exponent of $$x$$ to be 1012:

$$n b + 253 c = 1012$$

and the constraint $$a + b + c = 10$$ implies $$b + c \leq 10$$ since $$a = 10 - b - c \geq 0$$.

Given $$n \leq 22$$, we solve for non-negative integers $$b$$ and $$c$$ such that $$n b + 253 c = 1012$$ and $$b + c \leq 10$$. Rearranging for $$b$$:

$$b = \frac{1012 - 253 c}{n}$$

Since $$b$$ must be a non-negative integer, $$1012 - 253 c$$ must be divisible by $$n$$ and non-negative. Also, $$253 c \leq 1012$$ implies $$c \leq \frac{1012}{253} \approx 4.00$$, so $$c$$ can be 0, 1, 2, 3, or 4.

We check each possible $$c$$:

• For $$c = 0$$: $$b = \frac{1012}{n}$$. Then $$b + c = b \leq 10$$ implies $$\frac{1012}{n} \leq 10$$, so $$n \geq 101.2$$. But $$n \leq 22$$, contradiction. No solution.
• For $$c = 1$$: $$b = \frac{759}{n}$$. Then $$b + c = b + 1 \leq 10$$ implies $$b \leq 9$$, so $$\frac{759}{n} \leq 9$$ gives $$n \geq 84.333$$. Factorizing 759 = 3 × 11 × 23, divisors ≤22 are 1, 3, 11. But $$n \geq 84.333 > 22$$, no divisor satisfies. No solution.
• For $$c = 2$$: $$b = \frac{506}{n}$$. Then $$b + c = b + 2 \leq 10$$ implies $$b \leq 8$$, so $$\frac{506}{n} \leq 8$$ gives $$n \geq 63.25$$. Factorizing 506 = 2 × 11 × 23, divisors ≤22 are 1, 2, 11, 22. But $$n \geq 63.25 > 22$$, no divisor satisfies. No solution.
• For $$c = 3$$: $$b = \frac{253}{n}$$. Then $$b + c = b + 3 \leq 10$$ implies $$b \leq 7$$, so $$\frac{253}{n} \leq 7$$ gives $$n \geq 36.142$$. Factorizing 253 = 11 × 23, divisors ≤22 are 1, 11. But $$n \geq 36.142 > 22$$, no divisor satisfies. No solution.
• For $$c = 4$$: $$b = \frac{0}{n} = 0$$. Then $$b + c = 0 + 4 = 4 \leq 10$$, and $$a = 10 - 0 - 4 = 6$$. The exponent is $$n \cdot 0 + 253 \cdot 4 = 1012$$, which matches. This works for any $$n$$.

No other $$c$$ yields a solution given $$n \leq 22$$. Thus, the only solution is $$a = 6$$, $$b = 0$$, $$c = 4$$.

The coefficient is the multinomial coefficient:

$$\binom{10}{6, 0, 4} = \frac{10!}{6! 0! 4!}$$

Since $$0! = 1$$, this simplifies to:

$$\frac{10!}{6! 4!} = \binom{10}{4}$$

as $$\binom{10}{4} = \frac{10!}{4! 6!}$$.

Comparing with the options:

• A. $$^{253}C_4$$
• B. $$^{10}C_4$$
• C. $$4n$$
• D. $$1$$

The coefficient $$\binom{10}{4}$$ corresponds to option B.

Hence, the correct answer is Option B.

We have to find the coefficient of $$x^{50}$$ in the finite sum

$$S(x)= (1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+\,\ldots\,+x^{1000}.$$

The general (that is, the $$k$$-th) term of this sum is obtained by taking the exponent of $$x$$ equal to $$k$$ and the exponent of $$(1+x)$$ equal to $$1000-k$$. Thus the $$k$$-th term is

$$T_k \;=\;x^{k}\,(1+x)^{1000-k},\qquad k=0,1,2,\ldots,1000.$$

Inside this term the binomial expansion formula

$$ (1+x)^n \;=\;\sum_{r=0}^{n} \binom{n}{r}\,x^{\,r} $$

tells us that the power of $$x$$ that finally appears is $$k+r$$, and its coefficient there is $$\binom{1000-k}{r}.$$

We want the overall power to be exactly $$50$$, so we impose

$$k+r = 50.$$

Since $$k$$ runs from $$0$$ up to $$1000$$, only those $$k$$ with $$k\le 50$$ can possibly satisfy this equality (otherwise $$k+r$$ would exceed $$50$$ because $$r\ge 0$$). Therefore we let

$$k = 0,1,2,\ldots,50.$$

For a fixed $$k$$ in this range the corresponding $$r$$ is

$$r = 50-k,$$ so the contribution of the $$k$$-th term to the coefficient of $$x^{50}$$ is

$$\binom{1000-k}{\,50-k\,}.$$

Hence the required coefficient, call it $$C,$$ is the sum

$$ C \;=\;\sum_{k=0}^{50}\binom{1000-k}{\,50-k\,}. $$

It is convenient to reverse the index so that the upper number inside the binomial sign increases with the index. Put

$$r = 50-k \quad\Longrightarrow\quad k = 50-r.$$

When $$k$$ goes from $$0$$ to $$50$$, the new index $$r$$ goes from $$50$$ down to $$0$$, i.e. from $$0$$ to $$50$$ after re-ordering. Substituting $$k=50-r$$ inside the binomial coefficient gives

$$ \binom{1000-(50-r)}{\,r\,}\;=\;\binom{950+r}{\,r\,}. $$

Therefore

$$ C \;=\;\sum_{r=0}^{50}\binom{950+r}{\,r\,}. $$

Now we invoke a standard “hockey-stick” identity of binomial coefficients:

$$ \sum_{r=0}^{n}\binom{m+r}{\,r\,}\;=\;\binom{m+n+1}{\,n\,}. $$

Here $$m=950$$ and $$n=50,$$ so

$$ C \;=\;\binom{950+50+1}{\,50\,}\;=\;\binom{1001}{\,50\,}. $$

Writing this binomial coefficient in factorial form (using $$\displaystyle\binom{N}{k}=\frac{N!}{k!\,(N-k)!}$$) we obtain

$$ C \;=\;\frac{1001!}{50!\,951!}. $$

This expression exactly matches Option D in the list provided.

Hence, the correct answer is Option D.

In the binomial expansion of $$(a + b)^n$$, the general term $$T_{r+1}$$ is given by:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

The $$(r+1)^{th}$$ term is $$T_{r+1} = \binom{55}{r} (2)^{55-r} \left(\frac{x}{3}\right)^r = \left[ \binom{55}{r} 2^{55-r} \left(\frac{1}{3}\right)^r \right] x^r$$

The coefficient of the power $$x^r$$ is $$C_r = \binom{55}{r} 2^{55-r} 3^{-r}$$

We are looking for two consecutive terms with equal coefficients. Let these be the coefficients of $$x^r$$ and $$x^{r+1}$$. We set $$C_r = C_{r+1}$$:

$$\binom{55}{r} 2^{55-r} 3^{-r} = \binom{55}{r+1} 2^{55-(r+1)} 3^{-(r+1)}$$

$$\binom{55}{r} \cdot 2 = \binom{55}{r+1} \cdot \frac{1}{3}$$

$$6 \cdot \binom{55}{r} = \binom{55}{r+1}$$

Using the property $$\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1}$$:

$$6 = \frac{55 - r}{r + 1}$$

$$6(r + 1) = 55 - r$$

$$6r + 6 = 55 - r$$

$$7r = 49 \implies \mathbf{r = 7}$$

Since $$r = 7$$, the powers of $$x$$ with equal coefficients are $$x^7$$ and $$x^8$$:

The term containing $$x^7$$ is the $$(7+1)^{th} = \mathbf{8^{th}}$$ term.

The term containing $$x^8$$ is the $$(8+1)^{th} = \mathbf{9^{th}}$$ term.

We are given the expression $$(1+x)^{101}(1-x+x^2)^{100}$$ and need to find the number of distinct powers of $$x$$ in its expansion.

First, recognize that $$1 - x + x^2$$ can be rewritten using the identity $$1 + x^3 = (1 + x)(1 - x + x^2)$$. Solving for $$1 - x + x^2$$, we get:

$$1 - x + x^2 = \frac{1 + x^3}{1 + x}$$

for $$x \neq -1$$. Substituting this into the original expression:

$$(1+x)^{101} \cdot (1-x+x^2)^{100} = (1+x)^{101} \cdot \left( \frac{1 + x^3}{1 + x} \right)^{100} = (1+x)^{101} \cdot \frac{(1 + x^3)^{100}}{(1 + x)^{100}}$$

Simplify the exponents:

$$= (1+x)^{101-100} \cdot (1 + x^3)^{100} = (1+x) \cdot (1 + x^3)^{100}$$

So, the expression simplifies to $$(1+x)(1 + x^3)^{100}$$.

Now, expand $$(1 + x^3)^{100}$$ using the binomial theorem:

$$(1 + x^3)^{100} = \sum_{k=0}^{100} \binom{100}{k} (x^3)^k = \sum_{k=0}^{100} \binom{100}{k} x^{3k}$$

Multiply this by $$(1 + x)$$:

$$(1+x) \cdot \sum_{k=0}^{100} \binom{100}{k} x^{3k} = \sum_{k=0}^{100} \binom{100}{k} x^{3k} + x \cdot \sum_{k=0}^{100} \binom{100}{k} x^{3k} = \sum_{k=0}^{100} \binom{100}{k} x^{3k} + \sum_{k=0}^{100} \binom{100}{k} x^{3k+1}$$

This gives two separate sums:

1. The first sum: $$\sum_{k=0}^{100} \binom{100}{k} x^{3k}$$ has exponents $$3k$$ for $$k = 0, 1, 2, \ldots, 100$$. This produces exponents: $$0, 3, 6, \ldots, 300$$.
2. The second sum: $$\sum_{k=0}^{100} \binom{100}{k} x^{3k+1}$$ has exponents $$3k+1$$ for $$k = 0, 1, 2, \ldots, 100$$. This produces exponents: $$1, 4, 7, \ldots, 301$$.

Now, count the number of terms in each sum:

• For the first sum, exponents form an arithmetic sequence with first term 0, common difference 3, and last term 300. The number of terms is $$\frac{300 - 0}{3} + 1 = 100 + 1 = 101$$.
• For the second sum, exponents form an arithmetic sequence with first term 1, common difference 3, and last term 301. The number of terms is $$\frac{301 - 1}{3} + 1 = \frac{300}{3} + 1 = 100 + 1 = 101$$.

To find the total number of distinct powers, check if there is any overlap between the exponents of the two sums. Suppose an exponent $$m$$ appears in both sums. Then:

$$m = 3a \quad$$ for some integer $$a \quad$$ (from the first sum)

$$m = 3b + 1 \quad$$ for some integer $$b \quad$$ (from the second sum)

Equating these:

$$3a = 3b + 1 \implies 3a - 3b = 1 \implies 3(a - b) = 1$$

This equation has no integer solutions because 3 times an integer cannot equal 1. Therefore, there are no common exponents between the two sums.

Since there is no overlap, the total number of distinct exponents is the sum of the terms from both sums:

$$101 + 101 = 202$$

Hence, the number of distinct powers of $$x$$ is 202.

So, the answer is 202.

The binomial expansion of $$\left(2^{\frac{1}{2}} + 3^{\frac{1}{5}}\right)^{10}$$ has a general term given by the formula for the $$(r+1)$$-th term: $$T_{r+1} = \binom{10}{r} \left(2^{\frac{1}{2}}\right)^{10-r} \left(3^{\frac{1}{5}}\right)^r$$.

Simplifying the exponents, we get:

$$T_{r+1} = \binom{10}{r} 2^{\frac{10-r}{2}} \cdot 3^{\frac{r}{5}}$$

For a term to be rational, both exponents must be integers. This is because integer powers of integers yield rational numbers (specifically integers when the base is an integer and the exponent is non-negative). Therefore, we require:

1. $$\frac{10-r}{2}$$ must be an integer.
2. $$\frac{r}{5}$$ must be an integer.

Set $$\frac{r}{5} = m$$, where $$m$$ is an integer. Then $$r = 5m$$. Since $$r$$ must be between 0 and 10 inclusive, we have:

$$0 \leq 5m \leq 10 \implies 0 \leq m \leq 2$$

So $$m$$ can be 0, 1, or 2, giving $$r = 0, 5,$$ or $$10$$.

Now check the condition $$\frac{10-r}{2}$$ is an integer for each $$r$$:

• For $$r = 0$$: $$\frac{10-0}{2} = \frac{10}{2} = 5$$, which is an integer.
• For $$r = 5$$: $$\frac{10-5}{2} = \frac{5}{2} = 2.5$$, which is not an integer.
• For $$r = 10$$: $$\frac{10-10}{2} = \frac{0}{2} = 0$$, which is an integer.

Thus, only $$r = 0$$ and $$r = 10$$ yield rational terms. The term for $$r = 5$$ is irrational because $$2^{\frac{5}{2}} = \sqrt{32}$$, which is irrational.

Now compute the rational terms:

• For $$r = 0$$: $$T_1 = \binom{10}{0} \cdot 2^{\frac{10-0}{2}} \cdot 3^{\frac{0}{5}} = 1 \cdot 2^5 \cdot 3^0 = 1 \cdot 32 \cdot 1 = 32$$.
• For $$r = 10$$: $$T_{11} = \binom{10}{10} \cdot 2^{\frac{10-10}{2}} \cdot 3^{\frac{10}{5}} = 1 \cdot 2^0 \cdot 3^2 = 1 \cdot 1 \cdot 9 = 9$$.

The sum of the rational terms is $$32 + 9 = 41$$.

Hence, the correct answer is Option D.

The expansion of $$(1 + x)^{2n}$$ is given by the binomial theorem, and the general term is $$T_{k+1} = \binom{2n}{k} x^k$$, where the coefficient of $$x^k$$ is $$\binom{2n}{k}$$. According to the problem, the coefficients of $$x^{3r}$$ and $$x^{r+2}$$ are equal, so we have:

$$\binom{2n}{3r} = \binom{2n}{r+2}$$

We know that $$\binom{2n}{a} = \binom{2n}{b}$$ implies either $$a = b$$ or $$a + b = 2n$$. Therefore, we consider two cases.

Case 1: $$3r = r + 2$$

Solving this equation:

$$3r - r = 2$$

$$2r = 2$$

$$r = 1$$

But the problem specifies that $$r > 1$$, so $$r = 1$$ is invalid. This case is discarded.

Case 2: $$3r + (r + 2) = 2n$$

Solving this equation:

$$3r + r + 2 = 2n$$

$$4r + 2 = 2n$$

Divide both sides by 2:

$$2r + 1 = n$$

So, $$n = 2r + 1$$.

Now, we verify the conditions. Given $$r > 1$$, $$n = 2r + 1 > 2(1) + 1 = 3 > 2$$, so $$n > 2$$ is satisfied. Also, the indices $$3r$$ and $$r+2$$ must be within the range $$[0, 2n]$$. With $$n = 2r + 1$$, $$2n = 4r + 2$$.

Check $$3r$$: since $$r > 1$$, $$3r > 0$$ and $$3r \leq 4r + 2$$ (as $$3r \leq 4r + 2$$ simplifies to $$0 \leq r + 2$$, true for $$r > 1$$).

Check $$r+2$$: $$r+2 \leq 4r + 2$$ simplifies to $$0 \leq 3r$$, true for $$r > 1$$.

Additionally, $$3r \neq r+2$$ for $$r > 1$$ (since $$3r = r+2$$ gives $$r=1$$, excluded), so Case 2 applies.

Now, comparing with the options:

A. $$2r + 1$$

B. $$2r - 1$$

C. $$3r$$

D. $$r + 1$$

Option A matches $$n = 2r + 1$$.

To verify, let $$r = 2$$ (since $$r > 1$$), then $$n = 2(2) + 1 = 5$$, so $$2n = 10$$. Coefficient of $$x^{3r} = x^6$$ is $$\binom{10}{6}$$, and coefficient of $$x^{r+2} = x^4$$ is $$\binom{10}{4}$$. Since $$\binom{10}{6} = \binom{10}{4}$$ (as $$\binom{10}{6} = \binom{10}{10-6} = \binom{10}{4}$$), they are equal. Other options do not satisfy for $$r = 2$$.

Hence, the correct answer is Option A.

We begin with the binomial expansion formula. For any real numbers $$a$$ and $$b$$ and a positive integer $$n$$, the general term (also called the $$k^{\text{th}}$$ or $$(k+1)^{\text{th}}$$ term, depending on counting from zero) in $$(a+b)^n$$ is

$$T_{k+1}= {^nC_k}\, a^{\,n-k}\, b^{\,k},$$

where $$k = 0,1,2,\dots ,n$$ and $${^nC_k}=\dfrac{n!}{k!(n-k)!}$$.

In the given expression we have

$$a=\dfrac{3}{\sqrt[3]{84}},\qquad b=\sqrt{3}\,\ln x,\qquad n=9.$$

The problem refers to the 7th term, so we set $$k=6$$ (because $$T_{k+1}=T_7$$). Substituting these values into the general term formula, we get

$$T_7 = {^9C_6}\left(\dfrac{3}{\sqrt[3]{84}}\right)^{9-6}\bigl(\sqrt{3}\,\ln x\bigr)^{6}.$$

First, evaluate the binomial coefficient:

$$ {^9C_6}= \dfrac{9!}{6!\,3!}=84.$$

Next, handle the power of the first part:

$$\left(\dfrac{3}{\sqrt[3]{84}}\right)^{9-6}= \left(\dfrac{3}{\sqrt[3]{84}}\right)^{3} =\dfrac{3^{3}}{\bigl(\sqrt[3]{84}\bigr)^{3}} =\dfrac{27}{84} =\dfrac{9}{28}.$$

Putting these together, we obtain

$$T_7 = 84 \times \dfrac{9}{28} \times (\sqrt{3}\,\ln x)^{6}.$$

Simplify the numerical coefficient:

$$84 \times \dfrac{9}{28}= \bigl(\tfrac{84}{28}\bigr)\times 9 = 3 \times 9 = 27.$$

So

$$T_7 = 27\bigl(\sqrt{3}\,\ln x\bigr)^6.$$

According to the question, this 7th term equals $$729$$, hence

$$27\bigl(\sqrt{3}\,\ln x\bigr)^6 = 729.$$

Divide both sides by $$27$$:

$$(\sqrt{3}\,\ln x)^{6} = \dfrac{729}{27}=27.$$

Notice that $$27 = 3^{3}.$$ Therefore

$$(\sqrt{3}\,\ln x)^{6} = 3^{3}.$$

To isolate the inner expression, take the sixth root of both sides. Recall that $$3^{3} = 3^{\frac{6}{2}} = (3^{\frac12})^{6},$$ so

$$\sqrt{3}\,\ln x = 3^{\frac12} = \sqrt{3}.$$

Now divide by $$\sqrt{3}$$:

$$\ln x = 1.$$

Finally, use the definition of the natural logarithm ($$\ln x = 1 \Longrightarrow x = e^{1}$$):

$$x = e.$$

Looking at the provided options, $$e$$ corresponds to Option 2. Hence, the correct answer is Option 2.

We are given the binomial expansion of $$\left(x^2 + \frac{2}{x}\right)^{15}$$. The general term in the expansion is given by the binomial theorem. For $$(a + b)^n$$, the $$(r+1)^{\text{th}}$$ term is $$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$. Here, $$a = x^2$$, $$b = \frac{2}{x}$$, and $$n = 15$$.

So, the general term is:

$$T_{r+1} = \binom{15}{r} (x^2)^{15-r} \left(\frac{2}{x}\right)^r$$

Simplify the exponents:

$$T_{r+1} = \binom{15}{r} x^{2(15-r)} \cdot \frac{2^r}{x^r} = \binom{15}{r} x^{30 - 2r} \cdot 2^r \cdot x^{-r} = \binom{15}{r} 2^r x^{30 - 3r}$$

The exponent of $$x$$ is $$30 - 3r$$.

First, we find the coefficient of $$x^{15}$$. Set the exponent equal to 15:

$$30 - 3r = 15$$

Solve for $$r$$:

$$30 - 15 = 3r \implies 15 = 3r \implies r = 5$$

Substitute $$r = 5$$ into the general term to get the coefficient:

$$\text{Coefficient of } x^{15} = \binom{15}{5} 2^5$$

Next, we find the term independent of $$x$$, which means the exponent of $$x$$ is 0. Set the exponent equal to 0:

$$30 - 3r = 0$$

Solve for $$r$$:

$$30 = 3r \implies r = 10$$

Substitute $$r = 10$$ into the general term to get the coefficient:

$$\text{Coefficient of constant term} = \binom{15}{10} 2^{10}$$

Note that $$\binom{15}{10} = \binom{15}{5}$$ because $$\binom{n}{r} = \binom{n}{n-r}$$. So, $$\binom{15}{10} = \binom{15}{5}$$.

The ratio required is the coefficient of $$x^{15}$$ to the coefficient of the constant term:

$$\text{Ratio} = \frac{\binom{15}{5} \cdot 2^5}{\binom{15}{10} \cdot 2^{10}} = \frac{\binom{15}{5} \cdot 32}{\binom{15}{5} \cdot 1024}$$

Since $$\binom{15}{5}$$ is common and non-zero, it cancels out:

$$\text{Ratio} = \frac{32}{1024}$$

Simplify the fraction by dividing both numerator and denominator by 32:

$$\frac{32 \div 32}{1024 \div 32} = \frac{1}{32}$$

Thus, the ratio is $$1 : 32$$.

Comparing with the options, this matches Option D.

Hence, the correct answer is Option D.

We begin with the given expression

$$\left(\dfrac{x+1}{x^{2/3}-x^{1/3}+1}-\dfrac{x-1}{x-x^{1/2}}\right)^{10}.$$

Our target is the term that does not contain $$x$$ after expanding the entire power.

First we simplify each fraction inside the large brackets.

For the first fraction, we put $$t=x^{1/3} \;(\text{so}\; x=t^{3})$$. Then

$$x^{2/3}=t^{2},\qquad x^{1/3}=t.$$

Hence

$$\dfrac{x+1}{x^{2/3}-x^{1/3}+1}=\dfrac{t^{3}+1}{t^{2}-t+1}.$$

Now we recall the algebraic identity $$a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2}).$$ Setting $$a=t,\;b=1$$ gives

$$t^{3}+1=(t+1)(t^{2}-t+1).$$

Substituting into the numerator, we get

$$\dfrac{(t+1)(t^{2}-t+1)}{t^{2}-t+1}=t+1.$$

Undoing the substitution $$t=x^{1/3},$$ this simplifies to

$$x^{1/3}+1.$$

Next we simplify the second fraction. Put $$s=x^{1/2}\;(\text{so}\;x=s^{2}).$$ Then

$$x=s^{2},\qquad x^{1/2}=s.$$

Thus

$$\dfrac{x-1}{x-x^{1/2}}=\dfrac{s^{2}-1}{s^{2}-s}.$$

Factorising numerator and denominator, we use $$a^{2}-b^{2}=(a-b)(a+b)$$ to write

$$s^{2}-1=(s-1)(s+1),\qquad s^{2}-s=s(s-1).$$

Hence

$$\dfrac{(s-1)(s+1)}{s(s-1)}=\dfrac{s+1}{s}=1+\dfrac{1}{s}=1+s^{-1}.$$

Replacing $$s=x^{1/2}$$ gives

$$1+x^{-1/2}.$$

So the entire bracket now becomes

$$(x^{1/3}+1)-(1+x^{-1/2})=x^{1/3}-x^{-1/2}.$$

Therefore the whole expression is

$$(x^{1/3}-x^{-1/2})^{10}.$$

To expand, we use the Binomial Theorem, which states

$$ (u+v)^{n}=\sum_{k=0}^{n}\binom{n}{k}u^{\,n-k}v^{\,k}. $$

Here $$u=x^{1/3}$$ and $$v=-x^{-1/2},$$ with $$n=10.$$ The general term is

$$T_{k}=\binom{10}{k}(x^{1/3})^{\,10-k}\,(-x^{-1/2})^{\,k}.$$

Simplify the powers of $$x$$ in this term. The exponent of $$x$$ in $$T_{k}$$ is

$$\left(10-k\right)\!\left(\dfrac13\right)+k\!\left(-\dfrac12\right) =\dfrac{10-k}{3}-\dfrac{k}{2}.$$

For the term to be independent of $$x,$$ this exponent must be zero:

$$\dfrac{10-k}{3}-\dfrac{k}{2}=0.$$

Multiply by the common denominator $$6$$:

$$2(10-k)-3k=0.$$

So

$$20-2k-3k=0\quad\Longrightarrow\quad 20-5k=0\quad\Longrightarrow\quad k=4.$$

We substitute $$k=4$$ back into the general term:

$$T_{4}=\binom{10}{4}(x^{1/3})^{\,10-4}\,(-x^{-1/2})^{\,4} =\binom{10}{4}(x^{1/3})^{\,6}\,(-1)^{4}(x^{-1/2})^{\,4}.$$

Compute each factor:

$$(x^{1/3})^{6}=x^{6/3}=x^{2},\qquad (x^{-1/2})^{4}=x^{-2},\qquad (-1)^{4}=1.$$

The product $$x^{2}\cdot x^{-2}=x^{0}=1,$$ so all powers of $$x$$ cancel as required.

The coefficient is therefore

$$\binom{10}{4}=210.$$

Hence the term independent of $$x$$ is simply $$210.$$

Hence, the correct answer is Option A.

$$1 - x^3 = (1 - x)(1 + x + x^2)$$

$$(1 - x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1 - x)^{30 - 2r}$$

$$\left[(1 - x)(1 + x + x^2)\right]^{10} = (1 - x)^{10}(1 + x + x^2)^{10}$$

$$(1 - x)^{10}(1 + x + x^2)^{10} = \sum_{r=0}^{10} a_r x^r (1 - x)^{30 - 2r}$$

$$(1 + x + x^2)^{10} = \sum_{r=0}^{10} a_r x^r (1 - x)^{20 - 2r}$$

$$\frac{(1 + x + x^2)^{10}}{(1 - x)^{20}} = \sum_{r=0}^{10} a_r \frac{x^r}{(1 - x)^{2r}}$$

$$\left[ \frac{1 + x + x^2}{(1 - x)^2} \right]^{10} = \sum_{r=0}^{10} a_r \left[ \frac{x}{(1 - x)^2} \right]^r$$

$$\frac{1 + x + x^2}{(1 - x)^2} = \frac{(1 - 2x + x^2) + 3x}{(1 - x)^2} = \frac{(1 - x)^2 + 3x}{(1 - x)^2} = 1 + \frac{3x}{(1 - x)^2} = 1 + 3y$$

Let $$y = \frac{x}{(1 - x)^2}$$

$$(1 + 3y)^{10} = \sum_{r=0}^{10} a_r y^r$$

$$(1 + 3y)^{10} = \sum_{r=0}^{10} \binom{10}{r} (3y)^r = \sum_{r=0}^{10} \binom{10}{r} 3^r y^r$$

$$a_r = \binom{10}{r} 3^r$$

For $$r = 9$$: $$a_9 = \binom{10}{9} 3^9 = 10 \cdot 3^9$$

For $$r = 10$$: $$a_{10} = \binom{10}{10} 3^{10} = 1 \cdot 3^{10}$$

$$\frac{9a_9}{a_{10}} = \frac{9 \cdot (10 \cdot 3^9)}{3^{10}}$$

$$\frac{9a_9}{a_{10}} = \frac{10 \cdot 3^{11}}{3^{10}} = 10 \cdot 3 = 30$$

To solve this, we use the general term of the binomial expansion $$(a+b)^n$$, which is $$T_{r+1} = {}^nC_r \cdot a^{n-r} \cdot b^r$$.

For the expansion $$(\frac{1}{x^3} - x^4)^n$$:

$$T_{r+1} = {}^nC_r (x^{-3})^{n-r} (-x^4)^r = {}^nC_r (-1)^r x^{-3n + 3r + 4r} = {}^nC_r (-1)^r x^{7r - 3n}$$

• For $$x^7$$: $$7r_1 - 3n = 7 \implies 7r_1 = 3n + 7 \implies r_1 = \frac{3n+7}{7}$$
• For $$x^{14}$$: $$7r_2 - 3n = 14 \implies 7r_2 = 3n + 14 \implies r_2 = \frac{3n+14}{7}$$

Notice that $$r_2 = r_1 + 1$$.

The coefficients are $${}^nC_{r_1}(-1)^{r_1}$$ and $${}^nC_{r_2}(-1)^{r_2}$$. Given their sum is zero:

$${}^nC_{r_1}(-1)^{r_1} + {}^nC_{r_1+1}(-1)^{r_1+1} = 0$$

Divide by $$(-1)^{r_1}$$:

$${}^nC_{r_1} - {}^nC_{r_1+1} = 0 \implies {}^nC_{r_1} = {}^nC_{r_1+1}$$

In binomial coefficients, $${}^nC_x = {}^nC_y$$ implies either $$x = y$$ (not possible here) or $$x + y = n$$.

$$r_1 + (r_1 + 1) = n$$

$$2r_1 + 1 = n$$

Substitute $$r_1 = \frac{3n+7}{7}$$:

$$2\left(\frac{3n+7}{7}\right) + 1 = n$$

$$6n + 14 + 7 = 7n$$

$$21 = n$$

Value of $$n$$ is 21.

We need to find the least value of $$n$$ for which the binomial expansion of $$(\sqrt[3]{7} + \sqrt[12]{11})^n$$ has 183 integral terms.

The general term is: $$T_{r+1} = \binom{n}{r}(\sqrt[3]{7})^{n-r}(\sqrt[12]{11})^r = \binom{n}{r} \cdot 7^{(n-r)/3} \cdot 11^{r/12}$$

For $$T_{r+1}$$ to be an integer, the exponents of 7 and 11 must be non-negative integers:

(i) $$\frac{n-r}{3}$$ is a non-negative integer, i.e., $$n - r \equiv 0 \pmod{3}$$, i.e., $$r \equiv n \pmod{3}$$

(ii) $$\frac{r}{12}$$ is a non-negative integer, i.e., $$r \equiv 0 \pmod{12}$$

Since $$r$$ must be a multiple of 12, and 12 is a multiple of 3, condition (i) requires $$n \equiv 0 \pmod{3}$$.

When $$n$$ is a multiple of 3, both conditions are satisfied whenever $$r = 0, 12, 24, \ldots$$ with $$r \leq n$$.

The number of valid values of $$r$$ is $$\lfloor n/12 \rfloor + 1 = 183$$.

So $$\lfloor n/12 \rfloor = 182$$, which means $$2184 \leq n < 2196$$.

The least such $$n$$ that is also a multiple of 3 is $$n = 2184$$ (since $$2184 = 3 \times 728$$).

The correct answer is Option 1: 2184.

Expanding each binomial to second order, we have:

$$(1+x)^p \approx 1 + px + \frac{p(p-1)}{2}x^2 + \ldots$$

$$(1-x)^q \approx 1 - qx + \frac{q(q-1)}{2}x^2 + \ldots$$

Multiplying these series, the coefficient of $$x$$ is

$$p - q = 1 \quad\text{(i)}$$

and the coefficient of $$x^2$$ is

$$\frac{p(p-1)}{2} - pq + \frac{q(q-1)}{2} = -2 \quad\text{(ii)}$$

Multiplying equation (ii) by 2 and simplifying gives:

$$p^2 - p - 2pq + q^2 - q = -4,$$

which can be written as

$$(p - q)^2 - (p + q) = -4.$$

Since (i) implies $$(p - q)^2 = 1,$$ substituting into the last equation yields

$$1 - (p + q) = -4,$$

$$p + q = 5 \quad\text{(iii)}.$$

Solving the system $$p - q = 1$$ and $$p + q = 5$$ gives $$p = 3$$ and $$q = 2$$.

$$p^2 + q^2 = 9 + 4 = 13.$$

The correct answer is Option 2: 13.

Let's first simplify the ratio of the combinations $$\frac{^{30}C_{r}}{^{30}C_{r-1}}$$.
Using the identity $$\frac{^{n}C_{r}}{^{n}C_{r-1}} = \frac{n-r+1}{r}$$, we can substitute $$n = 30$$:

$$\frac{^{30}C_{r}}{^{30}C_{r-1}} = \frac{30-r+1}{r} = \frac{31-r}{r}$$

Now, substitute this back into our general term inside the summation:

$$\frac{r^{2} \left(^{30}C_{r}\right)^{2}}{^{30}C_{r-1}} = r^2 \cdot \left(^{30}C_{r}\right) \cdot \left(\frac{^{30}C_{r}}{^{30}C_{r-1}}\right)$$ $$= r^2 \cdot \left(^{30}C_{r}\right) \cdot \left(\frac{31-r}{r}\right)$$ $$= r(31-r) \cdot ^{30}C_{r}$$ $$= (31r - r^2) \cdot ^{30}C_{r}$$

$$\sum_{r=1}^{30} (31r - r^2) \cdot ^{30}C_{r} = 31 \sum_{r=1}^{30} r \cdot ^{30}C_{r} - \sum_{r=1}^{30} r^2 \cdot ^{30}C_{r}$$

1. First Sum:
   $$\sum_{r=1}^{n} r \cdot ^{n}C_{r} = n \cdot 2^{n-1}$$

   For $$n = 30$$:
   $$\sum_{r=1}^{30} r \cdot ^{30}C_{r} = 30 \cdot 2^{29}$$

2. Second Sum:
   $$\sum_{r=1}^{n} r^2 \cdot ^{n}C_{r} = n(n+1) \cdot 2^{n-2}$$

   For $$n = 30$$:
   $$\sum_{r=1}^{30} r^2 \cdot ^{30}C_{r} = 30(31) \cdot 2^{28} = 30 \cdot 31 \cdot \frac{2^{29}}{2} = 15 \cdot 31 \cdot 2^{29}$$

Substitute these values back into our equation:

$$\text{Sum} = 31 \left(30 \cdot 2^{29}\right) - \left(15 \cdot 31 \cdot 2^{29}\right)$$
Factor out $$31 \cdot 2^{29}$$:

$$\text{Sum} = 31 \cdot 2^{29} \cdot (30 - 15)$$ $$\text{Sum} = 31 \cdot 2^{29} \cdot 15$$ $$\text{Sum} = (31 \times 15) \times 2^{29}$$ $$\text{Sum} = 465 \times 2^{29}$$

Comparing this with the given right-hand side ($$\alpha \times 2^{29}$$):
$$\alpha = 465$$

Let $$(1 + x + x^2)^{10} = \displaystyle\sum_{r=0}^{20} a_r x^r$$.

Substituting $$x = 1$$: $$(1 + 1 + 1)^{10} = 3^{10} = \displaystyle\sum_{r=0}^{20} a_r$$.

Substituting $$x = -1$$: $$(1 - 1 + 1)^{10} = 1 = \displaystyle\sum_{r=0}^{20} a_r(-1)^r$$.

Subtracting: $$3^{10} - 1 = 2(a_1 + a_3 + a_5 + \ldots + a_{19})$$.

So $$a_1 + a_3 + a_5 + \ldots + a_{19} = \dfrac{3^{10} - 1}{2} = \dfrac{59049 - 1}{2} = 29524$$.

Now we need $$a_2$$. The coefficient of $$x^2$$ in $$(1 + x + x^2)^{10}$$:

Using the multinomial expansion $$(1 + x + x^2)^{10} = \displaystyle\sum \dfrac{10!}{i!j!k!}$$ where $$i + j + k = 10$$ and $$j + 2k = 2$$.

Possible values: $$(j, k) = (0, 1)$$ giving $$i = 9$$, or $$(j, k) = (2, 0)$$ giving $$i = 8$$.

$$a_2 = \dfrac{10!}{9!0!1!} + \dfrac{10!}{8!2!0!} = 10 + 45 = 55$$

Now, $$(a_1 + a_3 + \ldots + a_{19}) - 11a_2 = 29524 - 11(55) = 29524 - 605 = 28919$$.

$$28919 = 121k$$, so $$k = \dfrac{28919}{121} = 239$$.

Hence, the correct answer is 239.

We need to find $$\sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2}$$.

Using the identity: $$\frac{{}^{n}C_{k}}{k+1} = \frac{{}^{n+1}C_{k+1}}{n+1}$$

Here $$n = 11$$ and $$k = 2r+1$$, so:

$$\frac{{}^{11}C_{2r+1}}{2r+2} = \frac{{}^{12}C_{2r+2}}{12}$$

Therefore:

$$\sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{1}{12} \sum_{r=0}^{5} {}^{12}C_{2r+2}$$

The sum $$\sum_{r=0}^{5} {}^{12}C_{2r+2} = {}^{12}C_2 + {}^{12}C_4 + {}^{12}C_6 + {}^{12}C_8 + {}^{12}C_{10} + {}^{12}C_{12}$$

We know that the sum of even-indexed binomial coefficients equals: $$\sum_{k=0}^{6} {}^{12}C_{2k} = 2^{11} = 2048$$

This sum includes $${}^{12}C_0 = 1$$. So:

$$\sum_{r=0}^{5} {}^{12}C_{2r+2} = 2048 - {}^{12}C_0 = 2048 - 1 = 2047$$

Therefore:

$$\sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{2047}{12}$$

Since $$2047 = 23 \times 89$$ and $$12 = 2^2 \times 3$$, they share no common factors, so $$\gcd(2047, 12) = 1$$.

Thus $$m = 2047$$ and $$n = 12$$.

$$m - n = 2047 - 12 = 2035$$

Find the value of $$\alpha$$

We use the binomial expansion of $$(1 + i\sqrt{3})^{12}$$:

$$(1 + i\sqrt{3})^{12} = \binom{12}{0} + \binom{12}{1}(i\sqrt{3}) + \binom{12}{2}(i\sqrt{3})^2 + \binom{12}{3}(i\sqrt{3})^3 + \dots$$

The imaginary part is:

$$\text{Im}[(1 + i\sqrt{3})^{12}] = \binom{12}{1}(\sqrt{3}) - \binom{12}{3}(\sqrt{3})^3 + \binom{12}{5}(\sqrt{3})^5 - \dots$$

Divide by $$\sqrt{3}$$:

$$\frac{\text{Im}[(1 + i\sqrt{3})^{12}]}{\sqrt{3}} = \binom{12}{1} - 3\binom{12}{3} + 3^2\binom{12}{5} - \dots = \sum_{r=1}^{6}(-3)^{r-1} \binom{12}{2r-1}$$

Using De Moivre's Theorem:

$$1 + i\sqrt{3} = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3})$$

$$(1 + i\sqrt{3})^{12} = 2^{12}(\cos 4\pi + i\sin 4\pi) = 4096(1 + 0i)$$

Since the imaginary part is $$0$$, the summation equals $$0$$.

Therefore, $$\alpha = 1 + 0 = \mathbf{1}$$.

Calculate the distance

The line is $$x - \sqrt{3}y + 1 = 0$$. Distance from $$(12, \sqrt{3})$$:

$$d = \frac{|12 - \sqrt{3}(\sqrt{3}) + 1|}{\sqrt{1^2 + (-\sqrt{3})^2}} = \frac{|12 - 3 + 1|}{\sqrt{4}} = \frac{10}{2} = \mathbf{5}$$

We need to find the sum of all rational terms in the expansion of $$(1 + \sqrt{2} + \sqrt{3})^6$$.

By the multinomial theorem:

$$(1 + \sqrt{2} + \sqrt{3})^6 = \sum_{a+b+c=6} \frac{6!}{a!\,b!\,c!} \cdot 1^a \cdot (\sqrt{2})^b \cdot (\sqrt{3})^c$$

$$= \sum_{a+b+c=6} \frac{6!}{a!\,b!\,c!} \cdot 2^{b/2} \cdot 3^{c/2}$$

A term is rational when $$2^{b/2} \cdot 3^{c/2}$$ is rational, which requires both $$b$$ and $$c$$ to be even.

Let $$b = 2p$$ and $$c = 2q$$ where $$p, q \geq 0$$ and $$a + 2p + 2q = 6$$, so $$a = 6 - 2p - 2q$$.

We need $$a \geq 0$$, so $$p + q \leq 3$$:

$$S = 1 + 30 + 45 + 60 + 540 + 135 + 8 + 180 + 270 + 27 = 1296$$

The answer is $$\boxed{1296}$$.

Write a general term for the given series.
For the term containing $${}^{20}C_r$$ (where $$4 \le r \le 20$$) we have

$$\text{term} = (-1)^r\,(r-2)(r-3)\,{}^{20}C_r$$
because the sign alternates starting with $$+$$ for $$r=4$$, and
$$(r-2)(r-3)=2 \cdot 1,\; 3 \cdot 2,\; 4 \cdot 3,\ldots$$ exactly as in the question.

Denote the required sum by $$S$$:

$$S=\sum_{r=4}^{20}(-1)^r\,(r-2)(r-3)\,{}^{20}C_r$$ $$-(1)$$

First evaluate the easier sum running from $$r=0$$ to $$20$$ and then subtract the unwanted first four terms.

Define

$$T=\sum_{r=0}^{20}(-1)^r\,(r-2)(r-3)\,{}^{20}C_r$$ $$-(2)$$

Expand $$(r-2)(r-3)$$ so that each part can be handled by standard binomial identities:

$$(r-2)(r-3)=r(r-1)-5r+6$$

Therefore

$$T=\sum_{r=0}^{20}(-1)^r r(r-1)\,{}^{20}C_r -5\sum_{r=0}^{20}(-1)^r r\,{}^{20}C_r +6\sum_{r=0}^{20}(-1)^r {}^{20}C_r$$ $$-(3)$$

Each of the three sums in $$(3)$$ can be evaluated by differentiating $$(1+x)^{20}$$ at $$x=-1$$.

1. $$\displaystyle \sum_{r=0}^{20}(-1)^r{}^{20}C_r=(1-1)^{20}=0$$

2. Differentiate once:
$$20(1+x)^{19}=\sum_{r=0}^{20}r\,{}^{20}C_r x^{\,r-1}$$
Multiply by $$x$$ and put $$x=-1$$:

$$\sum_{r=0}^{20}(-1)^r r\,{}^{20}C_r = -20(1-1)^{19}=0$$

3. Differentiate twice:
$$20\!\cdot\!19\,(1+x)^{18}=\sum_{r=0}^{20} r(r-1)\,{}^{20}C_r x^{\,r-2}$$
Multiply by $$x^2$$ and put $$x=-1$$:

$$\sum_{r=0}^{20}(-1)^r r(r-1)\,{}^{20}C_r = 20\!\cdot\!19\,(1-1)^{18}=0$$

Hence every sum in $$(3)$$ equals $$0$$, so

$$T=0$$ $$-(4)$$

We now remove the terms for $$r=0,1,2,3$$ to obtain $$S$$.

Compute the four omitted terms:

For $$r=0$$: $$(-1)^0(0-2)(0-3){}^{20}C_0 = 6$$

For $$r=1$$: $$(-1)^1(1-2)(1-3){}^{20}C_1 = -2 \times 20 = -40$$

For $$r=2$$: $$(-1)^2(0)(-1){}^{20}C_2 = 0$$

For $$r=3$$: $$(-1)^3(1)(0){}^{20}C_3 = 0$$

The total of the excluded terms is $$6-40=-34$$.

Because $$T=0$$ from $$(4)$$, we have

$$S = T - (\text{excluded terms}) = 0 - (-34) = 34$$

Thus, the value of the given series is $$34$$.

The expression $$(x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5$$ for $$x>1$$ simplifies by letting $$a = x$$ and $$b = \sqrt{x^3 - 1}$$, so it becomes $$(a + b)^5 + (a - b)^5$$. Since the binomial theorem gives $$(a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5a b^4 + b^5$$ and $$(a - b)^5 = a^5 - 5a^4b + 10a^3b^2 - 10a^2b^3 + 5a b^4 - b^5$$, adding these expansions eliminates all odd powers of $$b$$.

This gives $$(a + b)^5 + (a - b)^5 = 2a^5 + 20a^3b^2 + 10ab^4 = 2\bigl(a^5 + 10a^3b^2 + 5ab^4\bigr)\,. $$

Substituting back $$a = x$$ and $$b^2 = x^3 - 1$$ (so that $$b^4 = (x^3 - 1)^2 = x^6 - 2x^3 + 1$$) transforms the expression into $$2\bigl[x^5 + 10x^3(x^3 - 1) + 5x(x^6 - 2x^3 + 1)\bigr]\,. $$

Expanding inside the brackets yields $$x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x = 5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x$$, and multiplying by 2 gives $$10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x\,. $$

From this polynomial the coefficients are $$\alpha=10$$ for $$x^7$$, $$\beta=2$$ for $$x^5$$, $$\gamma=-20$$ for $$x^3$$, and $$\delta=10$$ for $$x$$. The conditions $$\alpha u + \beta v = 18$$ and $$\gamma u + \delta v = 20$$ thus become $$10u + 2v = 18$$ and $$-20u + 10v = 20$$. Dividing the first by 2 gives $$5u + v = 9$$, and dividing the second by 10 gives $$-2u + v = 2$$. Subtracting the second from the first yields $$7u = 7$$, so $$u = 1$$, and substituting back gives $$v = 4$$. Therefore $$u + v = 5$$.

The correct option is A. 5.

The binomial expansion $$(x + y)^{m}$$ contains $$(m + 1)$$ terms. The sum of all its coefficients is obtained by substituting $$x = 1,\, y = 1$$, giving $$2^{m}$$.

Hence, the arithmetic mean (A.M.) of the coefficients is $$\text{A.M.} \;=\; \frac{\text{sum of coefficients}}{\text{number of coefficients}} = \frac{2^{m}}{m + 1}$$.

In this problem, $$m = 2n - 3$$ and the given arithmetic mean equals $$16$$, so $$\frac{2^{\,2n - 3}}{(2n - 3) + 1} = 16.$$ Simplifying the denominator, $$2n - 3 + 1 = 2n - 2$$, we get $$\frac{2^{\,2n - 3}}{2n - 2} = 16.$$

Multiply both sides by $$2n - 2$$: $$2^{\,2n - 3} = 16\,(2n - 2).$$ Since $$16 = 2^{4}$$ and $$2n - 2 = 2(n - 1)$$, rewrite the right‐hand side: $$2^{\,2n - 3} = 2^{4}\,\bigl[\,2(n - 1)\bigr] = 2^{5}\,(n - 1).$$

Both sides are powers of $$2$$, so equate the exponents: $$2n - 3 = 5 + \log_{2}(n - 1).$$ This is possible only when $$n - 1$$ itself is a power of $$2$$, because the left side is an integer.

Write $$2^{\,2n - 3} = 2^{5}\,(n - 1) \;\Longrightarrow\; 2^{\,2n - 8} = n - 1.$$ Now test integer $$\,n \ge 2$$:

Case 1: $$n = 2 \;\Rightarrow\; 2^{\,2\cdot2 - 8} = 2^{-4} = \tfrac{1}{16} \neq 1.$$
Case 2: $$n = 3 \;\Rightarrow\; 2^{\,6 - 8} = 2^{-2} = \tfrac{1}{4} \neq 2.$$
Case 3: $$n = 4 \;\Rightarrow\; 2^{\,8 - 8} = 2^{0} = 1 \neq 3.$$
Case 4: $$n = 5 \;\Rightarrow\; 2^{\,10 - 8} = 2^{2} = 4 = 5 - 1.$$

Equality holds only for $$n = 5$$. No larger integer satisfies the equation because the left side then grows much faster than the right side. Thus, $$n = 5$$.

The coordinates of point $$P$$ are $$x_{P} = 2n - 1 = 2(5) - 1 = 9,$$ $$y_{P} = n^{2} - 4n = 5^{2} - 4\cdot5 = 25 - 20 = 5.$$ So $$P(9,\,5).$$

The line is $$x + y = 8$$, written in standard form as $$x + y - 8 = 0,$$ where $$a = 1,\, b = 1,\, c = -8.$$

Distance of $$P(x_{0},y_{0})$$ from $$ax + by + c = 0$$ is $$\text{Distance} = \frac{\lvert ax_{0} + by_{0} + c \rvert}{\sqrt{a^{2} + b^{2}}}.$$ Substitute $$x_{0} = 9,\, y_{0} = 5$$:

Numerator: $$\lvert 1\cdot9 + 1\cdot5 - 8 \rvert = \lvert 6 \rvert = 6.$$ Denominator: $$\sqrt{1^{2} + 1^{2}} = \sqrt{2}.$$

Therefore, $$\text{Distance} = \frac{6}{\sqrt{2}} = 3\sqrt{2}.$$

Hence the required distance is $$3\sqrt{2}$$, which corresponds to Option D.

This problem has two parts. We need to find $$p$$ (number of values of $$r$$) and $$q$$ (sum of rational terms), then compute $$p + q$$.

Part 1: Finding p

In the expansion of $$(a+b)^{12}$$, the general term is $$T_{k+1} = \binom{12}{k} a^{12-k} b^k$$.

The coefficients of $$T_r, T_{r+1}, T_{r+2}$$ are $$\binom{12}{r-1}, \binom{12}{r}, \binom{12}{r+1}$$ respectively.

For these to be in G.P., we need:

$$\binom{12}{r}^2 = \binom{12}{r-1} \cdot \binom{12}{r+1}$$

Using the identity $$\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$$:

$$\frac{\binom{12}{r}}{\binom{12}{r-1}} = \frac{13-r}{r}$$ and $$\frac{\binom{12}{r+1}}{\binom{12}{r}} = \frac{12-r}{r+1}$$

The G.P. condition gives: $$\frac{\binom{12}{r}}{\binom{12}{r-1}} = \frac{\binom{12}{r+1}}{\binom{12}{r}}$$

$$\frac{13-r}{r} = \frac{12-r}{r+1}$$

$$(13-r)(r+1) = r(12-r)$$

$$13r + 13 - r^2 - r = 12r - r^2$$

$$12r + 13 - r^2 = 12r - r^2$$

$$13 = 0$$

This is a contradiction! This means no three consecutive binomial coefficients of $$(a+b)^{12}$$ can form a G.P. Therefore $$p = 0$$.

Part 2: Finding q (sum of rational terms)

In the expansion of $$(3^{1/4} + 4^{1/3})^{12}$$, the general term is:

$$T_{k+1} = \binom{12}{k} (3^{1/4})^{12-k} (4^{1/3})^k = \binom{12}{k} \cdot 3^{(12-k)/4} \cdot 4^{k/3}$$

Since $$4 = 2^2$$, we have $$4^{k/3} = 2^{2k/3}$$.

For the term to be rational, we need both $$\frac{12-k}{4}$$ and $$\frac{k}{3}$$ to be non-negative integers.

From $$\frac{12-k}{4} \in \mathbb{Z}_{\geq 0}$$: $$k \equiv 0 \pmod{4}$$ and $$k \leq 12$$.

From $$\frac{k}{3} \in \mathbb{Z}_{\geq 0}$$: $$k \equiv 0 \pmod{3}$$.

So $$k$$ must be divisible by both 3 and 4, i.e., $$k \equiv 0 \pmod{12}$$.

For $$0 \leq k \leq 12$$: $$k = 0$$ or $$k = 12$$.

For $$k = 0$$: $$T_1 = \binom{12}{0} \cdot 3^{12/4} \cdot 4^0 = 1 \cdot 3^3 \cdot 1 = 27$$

For $$k = 12$$: $$T_{13} = \binom{12}{12} \cdot 3^0 \cdot 4^{12/3} = 1 \cdot 1 \cdot 4^4 = 256$$

Therefore $$q = 27 + 256 = 283$$.

Finally: $$p + q = 0 + 283 = 283$$, which matches Option A.

First write the given power in a compact form.
$$((64)^{(64)})^{(64)} = \left(64^{64}\right)^{64}$$

Using the law of exponents $$\left(a^{m}\right)^{n}=a^{mn}$$, we get
$$\left(64^{64}\right)^{64}=64^{64\times 64}=64^{4096}$$

We need the remainder of $$64^{4096}$$ when divided by 7. For remainders, replace each base by its least positive residue modulo 7:
$$64 \equiv 64-63 = 1 \pmod{7}$$ because $$7 \times 9 = 63$$.

Therefore
$$64^{4096}\equiv 1^{4096}\pmod{7}$$

Any power of 1 is still 1, so
$$1^{4096}=1$$

Hence the remainder when $$((64)^{(64)})^{(64)}$$ is divided by 7 equals $$1$$.

The correct option is Option B (1).

Write the binomial expansion of $$(2+\sqrt{3})^{8}$$:

$$ (2+\sqrt{3})^{8} = \sum_{k=0}^{8} \binom{8}{k}\,2^{\,8-k}\,(\sqrt{3})^{\,k} $$

The factor $$(\sqrt{3})^{\,k}$$ is rational only when $$k$$ is even, because

$$ (\sqrt{3})^{\,k} = \begin{cases} 3^{\,k/2}, & k \text{ even}\\[4pt] 3^{\,(k-1)/2}\sqrt{3}, & k \text{ odd} \end{cases} $$

Hence we keep only even values of $$k$$ (0, 2, 4, 6, 8) and evaluate each term.

Case 1: $$k = 0$$
$$ \binom{8}{0}\,2^{\,8}\,3^{\,0} = 1 \times 256 \times 1 = 256 $$

Case 2: $$k = 2$$
$$ \binom{8}{2}\,2^{\,6}\,3^{\,1} = 28 \times 64 \times 3 = 5376 $$

Case 3: $$k = 4$$
$$ \binom{8}{4}\,2^{\,4}\,3^{\,2} = 70 \times 16 \times 9 = 10080 $$

Case 4: $$k = 6$$
$$ \binom{8}{6}\,2^{\,2}\,3^{\,3} = 28 \times 4 \times 27 = 3024 $$

Case 5: $$k = 8$$
$$ \binom{8}{8}\,2^{\,0}\,3^{\,4} = 1 \times 1 \times 81 = 81 $$

Add the five rational contributions:

$$ 256 + 5376 + 10080 + 3024 + 81 = 18817 $$

The sum of all rational terms in the expansion of $$(2+\sqrt{3})^{8}$$ is $$18817$$.

Therefore, the correct option is Option D.