JEE Binomial Theorem Questions

Rewrite the binomial in the order that will make the calculation convenient (addition is commutative):
$$\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^n =\left(\dfrac{1}{\sqrt[3]{3}}+\sqrt[3]{2}\right)^n.$$

Let
$$a=\dfrac{1}{\sqrt[3]{3}}=3^{-1/3},\qquad b=\sqrt[3]{2}=2^{1/3}.$$

The general term in the expansion of $$(a+b)^n$$ is
$$T_{r+1} = {}^nC_r\,a^{\,n-r}\,b^{\,r},\qquad r=0,1,2,\ldots ,n.$$

15th term from the beginning
Here, $$r=14$$ (because the first term corresponds to $$r=0$$).
$$T_{15} = {}^nC_{14}\,a^{\,n-14}\,b^{\,14}.$$

15th term from the end
The $$k^{\text{th}}$$ term from the end is obtained by putting $$r = n-k+1.$$ For $$k=15$$ we get $$r = n-14.$$
$$T_{\text{end}} = {}^nC_{\,n-14}\,a^{\,14}\,b^{\,n-14}.$$ Because $$\binom{n}{n-14} = \binom{n}{14},$$ the binomial coefficients of the two terms are equal.

Given ratio
$$\dfrac{\text{15th term from end}}{\text{15th term from beginning}} = \dfrac{1}{6}.$$ Substituting the two terms (the coefficients cancel):
$$\dfrac{a^{14}\,b^{\,n-14}}{a^{\,n-14}\,b^{14}} = \dfrac{1}{6}.$$

Simplify the left-hand side:
$$a^{\,14-(n-14)}\,b^{\,(n-14)-14} = a^{\,28-n}\,b^{\,n-28}.$$ Insert $$a=3^{-1/3}$$ and $$b=2^{1/3}:$$
$$\bigl(3^{-1/3}\bigr)^{28-n}\, \bigl(2^{1/3}\bigr)^{n-28} = 3^{\,(28-n)(-1/3)}\, 2^{\,(n-28)/3} = 3^{\,(n-28)/3}\,2^{\,(n-28)/3} = 6^{\,(n-28)/3}.$$

Hence
$$6^{\,(n-28)/3} = \dfrac{1}{6} = 6^{-1}.$$ Equate the exponents (bases are equal and positive):
$$\dfrac{n-28}{3} = -1 \;\;\Longrightarrow\;\; n-28 = -3 \;\;\Longrightarrow\;\; n = 25.$$

Finally, compute $${}^nC_3$$:
$${}^{25}C_3 = \dfrac{25\times 24\times 23}{3\times 2\times 1} = 2300.$$

Therefore, $${}^nC_3 = 2300,$$ which is Option C.

The required sum is $$S = \displaystyle\sum_{r=1}^{15} r^{2}\,\binom{15}{r}$$.

Start from the binomial expansion: $$(1+x)^{n} = \displaystyle\sum_{r=0}^{n} \binom{n}{r}\,x^{r}$$.

Differentiating once with respect to $$x$$ gives
$$\frac{d}{dx}(1+x)^{n} = n(1+x)^{\,n-1} = \sum_{r=1}^{n} r\,\binom{n}{r}\,x^{\,r-1}$$.

Multiplying by $$x$$: $$n(1+x)^{\,n-1}\,x = \sum_{r=1}^{n} r\,\binom{n}{r}\,x^{\,r}$$.

Differentiating a second time:
$$\frac{d}{dx}\!\left[n(1+x)^{\,n-1}\,x\right] = n(n-1)(1+x)^{\,n-2}\,x + n(1+x)^{\,n-1} = \sum_{r=1}^{n} r^{2}\,\binom{n}{r}\,x^{\,r-1}$$.

Again multiply by $$x$$ to match the powers:
$$n(n-1)(1+x)^{\,n-2}\,x^{2} + n(1+x)^{\,n-1}\,x = \sum_{r=1}^{n} r^{2}\,\binom{n}{r}\,x^{\,r}$$.

Now put $$x = 1$$ (because $$2^{n} = (1+1)^{n}$$):
Left side becomes $$n(n-1)\,2^{\,n-2}\,(1)^{2} + n\,2^{\,n-1}\,(1)$$.
Hence
$$\sum_{r=1}^{n} r^{2}\,\binom{n}{r} = n(n-1)\,2^{\,n-2} + n\,2^{\,n-1}$$.

Simplify the right side:
$$n(n-1)\,2^{\,n-2} + n\,2^{\,n-1} = n\,2^{\,n-2}\,\big[(n-1) + 2\big] = n(n+1)\,2^{\,n-2}$$ $$-(1)$$.

For our problem, $$n = 15$$. Substituting in $$(1)$$:
$$S = 15\,(15+1)\,2^{\,15-2} = 15 \times 16 \times 2^{\,13}$$.

Compute the numerical coefficients:
$$15 \times 16 = 240 = 2^{4}\times 3 \times 5$$.

Therefore
$$S = 240 \times 2^{\,13} = \bigl(2^{4}\times 3 \times 5\bigr)\,2^{\,13} = 2^{\,17}\,3^{\,1}\,5^{\,1}$$.

The required exponents are $$m = 17,\; n = 1,\; k = 1$$.

Hence $$m + n + k = 17 + 1 + 1 = 19$$.

So the correct option is Option A (19).

1. Simplify the General Term

We begin by simplifying the fraction inside the summation:

$$\frac{10^{r+1}-1}{10^r} = \frac{10^{r+1}}{10^r} - \frac{1}{10^r} = 10 - 10^{-r}$$

Substituting this back into the sum, we can split it into two separate series:

$$S = \sum_{r=0}^{10} \left(10 - 10^{-r}\right) {}^{11}C_{r+1} = 10\sum_{r=0}^{10} {}^{11}C_{r+1} - \sum_{r=0}^{10} 10^{-r} \cdot {}^{11}C_{r+1}$$

To make things easier, let's use a change of variable. Let $$k = r + 1$$. As $$r$$ varies from $$0$$ to $$10$$, $$k$$ runs from $$1$$ to $$11$$:

$$
\begin{aligned}
S = 10\sum_{k=1}^{11} {}^{11}C_k - \sum_{k=1}^{11} 10^{-(k-1)} {}^{11}C_k && \text{(1)}
\end{aligned}
$$

2. Evaluate the First Summation

Using the standard binomial theorem identity $$\sum_{k=0}^{11} {}^{11}C_k = 2^{11}$$, we subtract the missing $$k=0$$ term ($${}^{11}C_0 = 1$$):

$$\sum_{k=1}^{11} {}^{11}C_k = 2^{11} - {}^{11}C_0 = 2^{11} - 1$$

Multiplying by the constant factor of $$10$$:

$$10\sum_{k=1}^{11} {}^{11}C_k = 10\left(2^{11} - 1\right) = 10(2048 - 1) = 20470 \tag{2}$$

3. Evaluate the Second Summation

We rewrite the exponent $$10^{-(k-1)}$$ as $$10 \cdot 10^{-k}$$:

$$\sum_{k=1}^{11} 10^{-(k-1)} {}^{11}C_k = 10\sum_{k=1}^{11} 10^{-k} {}^{11}C_k$$

Using the binomial expansion formula $$(1+x)^{11} = \sum_{k=0}^{11} {}^{11}C_k x^k$$ with $$x = \frac{1}{10}$$:

$$\sum_{k=1}^{11} {}^{11}C_k 10^{-k} = \left(1 + \frac{1}{10}\right)^{11} - {}^{11}C_0 = \left(\frac{11}{10}\right)^{11} - 1$$

Now, multiply this by $$10$$:

$$\sum_{k=1}^{11} 10^{-(k-1)} {}^{11}C_k = 10\left[\left(\frac{11}{10}\right)^{11} - 1\right] = 10 \cdot \frac{11^{11}}{10^{11}} - 10 = \frac{11^{11}}{10^{10}} - 10 \tag{3}$$

4. Combine and Solve for $$\alpha$$

Substitute equation (2) and equation (3) back into the primary expression for $$S$$ (Equation 1):

$$S = 20470 - \left( \frac{11^{11}}{10^{10}} - 10 \right)$$ $$S = 20470 + 10 - \frac{11^{11}}{10^{10}}$$  $$S = 20480 - \frac{11^{11}}{10^{10}}$$

To match the target format given in the problem, let's establish a common denominator of $$10^{10}$$:

$$S = \frac{20480 \cdot 10^{10} - 11^{11}}{10^{10}}$$

We are given that:

$$S = \frac{\alpha^{11} - 11^{11}}{10^{10}}$$

By comparing the numerators directly:

$$\alpha^{11} = 20480 \cdot 10^{10}$$

Since $$2048 = 2^{11}$$, we can rewrite $$20480$$ as $$2^{11} \cdot 10$$:

$$\alpha^{11} = (2^{11} \cdot 10) \cdot 10^{10}$$  $$\alpha^{11} = 2^{11} \cdot 10^{11}$$  $$\alpha^{11} = (2 \cdot 10)^{11} = 20^{11}$$

Taking the $$11\text{-th}$$ root on both sides:

$$\alpha = 20$$

Final Answer

Option D (20)

Identify coefficients

In $$(1+x)^{2n-1}$$:

• $$A = T_{30} \text{ coeff} = \binom{2n-1}{29}$$
• $$B = T_{12} \text{ coeff} = \binom{2n-1}{11}$$

Step 2: Use the ratio 2A = 5B

$$2 \cdot \frac{(2n-1)!}{29!(2n-30)!} = 5 \cdot \frac{(2n-1)!}{11!(2n-12)!}$$

Cancel $$(2n-1)!$$ and rearrange:

$$\frac{2}{5} = \frac{29!}{11!} \cdot \frac{(2n-30)!}{(2n-12)!}$$

$$\frac{2}{5} = \frac{29 \cdot 28 \cdot ... \cdot 12}{(2n-12) \cdot (2n-13) \cdot ... \cdot (2n-29)}$$

Alternatively, using the property $$\binom{n}{r} = \binom{n}{n-r}$$, if we test $$n=21$$:

$$2 \binom{41}{29} = 5 \binom{41}{11}$$

Since $$\binom{41}{29} = \binom{41}{41-29} = \binom{41}{12}$$:

$$2 \binom{41}{12} = 5 \binom{41}{11} \implies 2 \frac{41!}{12!29!} = 5 \frac{41!}{11!30!}$$

$$2/12 = 5/30 \implies 1/6 = 1/6$$. (Matches!)

Correct Option: C (21)

Let the general term in the binomial expansion of $$\left(5^{1/2}+7^{1/8}\right)^{1016}$$ be $$T_{k+1}$$, where $$k$$ ranges from $$0$$ to $$1016$$.

Using the Binomial Theorem,

$$T_{k+1} = \binom{1016}{k}\,\left(5^{1/2}\right)^{1016-k}\,\left(7^{1/8}\right)^{k}$$

Simplify the powers of $$5$$ and $$7$$:

$$T_{k+1} = \binom{1016}{k}\,5^{\frac{1016-k}{2}}\;7^{\frac{k}{8}}$$ $$-(1)$$

For $$T_{k+1}$$ to be an integer, the exponents of both $$5$$ and $$7$$ must be non-negative integers.

Condition 1 (for the power of 7)
The exponent $$\dfrac{k}{8}$$ must be an integer ⟹ $$k$$ must be a multiple of $$8$$.

Condition 2 (for the power of 5)
The exponent $$\dfrac{1016-k}{2}$$ must be an integer. Since $$1016$$ is even, $$\dfrac{1016-k}{2}$$ is an integer whenever $$k$$ is even.

If $$k$$ is a multiple of $$8$$, then $$k$$ is automatically even, so both conditions are satisfied simultaneously.

Hence, the permissible values of $$k$$ are all multiples of $$8$$ from $$0$$ up to $$1016$$ (inclusive):

$$k = 0,\,8,\,16,\,24,\,\dots,\,1016$$

The number of terms in this arithmetic sequence is

$$\text{Count} = \frac{1016}{8} + 1 = 127 + 1 = 128$$

Therefore, the expansion contains $$128$$ integral terms.

Option D is correct.

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.

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.

Coefficients of 5th, 6th and 7th terms of $$(1+x)^{n+4}$$ are in arithmetic progression.

Let N = n + 4 so that the coefficient of the rth term is $$\binom{N}{r-1}$$.

Hence the 5th term has coefficient $$\binom{N}{4}$$, the 6th term has coefficient $$\binom{N}{5}$$, and the 7th term has coefficient $$\binom{N}{6}$$.

Imposing the arithmetic progression condition yields $$2\binom{N}{5} = \binom{N}{4} + \binom{N}{6}$$. 

Using the identities $$\binom{N}{5} = \frac{N-4}{5}\binom{N}{4}$$ and $$\binom{N}{6} = \frac{N-5}{6}\binom{N}{5}$$ and dividing through by $$\binom{N}{4}$$ gives $$2\frac{N-4}{5} = 1 + \frac{(N-4)(N-5)}{30}$$.

Multiplying both sides by 30 leads to $$12(N-4) = 30 + (N-4)(N-5)$$. 

Setting $$m = N-4$$ transforms this into $$12m = 30 + m(m-1)$$, or $$m^2 - 13m + 30 = 0$$.

$$(m-3)(m-10) = 0$$ so that $$m = 3$$ or $$m = 10$$. Thus $$N = 7$$ or $$N = 14$$, giving $$n = 3$$ or $$n = 10$$. Excluding $$n = 10$$, we have $$n = 3$$ and hence $$N = 7$$.

In the expansion of $$(1+x)^7$$ the largest binomial coefficient is $$\binom{7}{3} = \binom{7}{4} = 35$$.

The correct answer is Option 3: 35.

Write the expression inside the tenth power as

$$\Delta = \frac{x+1}{x^{2/3}+1-x^{1/3}}-\frac{x+1}{x-x^{1/2}},\qquad x\gt 1$$

Step 1 (first fraction)
Factor the denominator as $$1-x^{1/3}+x^{2/3}=\dfrac{1+x}{1+x^{1/3}}$$ because
$$1-x^{1/3}+x^{2/3}=(x^{1/3})^{2}-x^{1/3}+1=\frac{x+1}{x^{1/3}+1}.$$
Hence

$$\frac{x+1}{x^{2/3}+1-x^{1/3}} =\frac{x+1}{\dfrac{x+1}{1+x^{1/3}}}=1+x^{1/3}.$$

Step 2 (second fraction)
Factor $$x$$ from the denominator:

$$\frac{x+1}{x-x^{1/2}} =\frac{x+1}{x\bigl(1-x^{-1/2}\bigr)} =\frac{1+x^{-1}}{1-x^{-1/2}}.$$ For $$x\gt 1$$ we have $$|x^{-1/2}|<1,$$ so use the geometric series $$\frac{1}{1-x^{-1/2}}=\sum_{k=0}^{\infty}x^{-k/2}.$$ Thus

$$\frac{x+1}{x-x^{1/2}} =(1+x^{-1})\sum_{k=0}^{\infty}x^{-k/2} =1+x^{-1/2}+2x^{-1}+2x^{-3/2}+2x^{-2}+\cdots\;.$$

Step 3 (series for Δ)
Subtract the series just obtained from $$1+x^{1/3}:$$

$$\Delta=(1+x^{1/3})- \bigl[1+x^{-1/2}+2x^{-1}+2x^{-3/2}+2x^{-2}+\cdots\bigr] =x^{1/3}-x^{-1/2}-2x^{-1}-2x^{-3/2}-2x^{-2}-\cdots\;.$$

Keep the powers of $$x$$ together with their coefficients:

Term T1: coefficient $$1$$, power $$x^{1/3}$$.
Term T2: coefficient $$-1$$, power $$x^{-1/2}$$.
Term T3: coefficient $$-2$$, power $$x^{-1}$$.
Term T4: coefficient $$-2$$, power $$x^{-3/2}$$, and so on.

Step 4 (condition for constant term in $$\Delta^{10}$$)
Choose $$a$$ copies of T1, $$b$$ copies of T2, $$c$$ copies of T3, $$d$$ copies of T4, … in the multinomial expansion, where

$$a+b+c+d+\cdots = 10\qquad -(1)$$ and the total exponent of $$x$$ must be zero:

$$\frac{1}{3}a-\frac{1}{2}b-1c-\frac{3}{2}d-\cdots = 0.$$

Multiply by $$6$$ to avoid fractions:

$$2a-3b-6c-9d-\cdots = 0\qquad -(2)$$

Because every negative coefficient in $$(2)$$ is a multiple of $$3,$$ the left-hand side can vanish only if $$2a$$ is also a multiple of $$3.$$
Hence $$a$$ itself must be a multiple of $$3,$$ so $$a=0,3,6,9$$ (but $$\le 10$$).

Case a = 0: the sum in $$(2)$$ is negative ⇒ impossible.

Case a = 3: $$2a=6.$$ Then $$(2)\Rightarrow 3b+6c+9d+\cdots = 6 \Rightarrow b+2c+3d+\cdots = 2.$$ Equation $$(1)$$ gives $$b+c+d+\cdots = 7,$$ impossible together.

Case a = 6: $$2a=12.$$ Then $$b+2c+3d+\cdots = 4\qquad -(3)$$ and $$b+c+d+\cdots = 4\qquad -(4).$$ Subtract $$(4)$$ from $$(3)$$ to get $$c+2d+3e+\cdots = 0,$$ which forces $$c=d=e=\cdots = 0.$$ Consequently $$b=4.$$ Thus the only viable solution is
$$a=6,\; b=4,\; c=d=e=\cdots =0.$$

Case a = 9: leads to contradictions similar to the case $$a=3$$ (no solution).

Therefore the constant term arises solely from choosing
6 factors of $$x^{1/3}$$ and 4 factors of $$-x^{-1/2}.$$

Step 5 (coefficient)
The multinomial coefficient is $$\dfrac{10!}{6!4!} = 210.$$ The numerical factors from the chosen terms are $$(1)^6(-1)^4 = 1.$$

Hence the term independent of $$x$$ in $$\Delta^{10}$$ is $$210.$$

Answer : Option A  - 210

We need to evaluate the finite sum
$$S=\sum_{r=1}^{9}\frac{\,r+3\,}{2^{r}}\;{}^{9}C_{r}$$
and express it in the form
$$S=\alpha\left(\frac{3}{2}\right)^{9}-\beta,\qquad\alpha,\beta\in\mathbb{N}.$$

Split $$S$$ into two simpler sums:
$$S=\sum_{r=1}^{9}\frac{r}{2^{r}}\;{}^{9}C_{r}+3\sum_{r=1}^{9}\frac{1}{2^{r}}\;{}^{9}C_{r}$$
Let
$$S_{1}=\sum_{r=1}^{9}\frac{r}{2^{r}}\;{}^{9}C_{r},\qquad S_{2}=\sum_{r=1}^{9}\frac{1}{2^{r}}\;{}^{9}C_{r}.$$

Case 1: Calculation of $$S_{1}$$
Start with the binomial expansion:
$$(1+x)^{9}=\sum_{r=0}^{9}{}^{9}C_{r}\,x^{r}.$$
Differentiate both sides with respect to $$x$$:
$$9(1+x)^{8}=\sum_{r=1}^{9}r\,{}^{9}C_{r}\,x^{\,r-1}.$$
Multiply by $$x$$ to match the power $$x^{r}$$:
$$9x(1+x)^{8}=\sum_{r=1}^{9}r\,{}^{9}C_{r}\,x^{\,r}.$$
Hence
$$S_{1}=\sum_{r=1}^{9}r\,{}^{9}C_{r}\left(\frac{1}{2}\right)^{r}=9\left(\frac{1}{2}\right)(1+\tfrac12)^{8}.$$
Now $$1+\tfrac12=\tfrac32$$, so
$$S_{1}=9\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)^{8} =\frac{9}{2}\left(\frac{3}{2}\right)^{8}.$$

Rewrite $$S_{1}$$ with the power $$\left(\tfrac32\right)^{9}$$:
$$\frac{9}{2}\left(\frac{3}{2}\right)^{8} =\frac{9}{2}\left(\frac{3}{2}\right)^{9}\!\!\left(\frac{2}{3}\right) =3\left(\frac{3}{2}\right)^{9}.$$

Case 2: Calculation of $$S_{2}$$
Again use the binomial expansion:
$$(1+x)^{9}=\sum_{r=0}^{9}{}^{9}C_{r}\,x^{r}.$$
Put $$x=\tfrac12$$ and subtract the $$r=0$$ term:
$$S_{2}=(1+\tfrac12)^{9}-1=\left(\frac{3}{2}\right)^{9}-1.$$

Combine the results
$$S=S_{1}+3S_{2} =3\left(\frac{3}{2}\right)^{9}+3\left[\left(\frac{3}{2}\right)^{9}-1\right]$$
$$\quad=3\left(\frac{3}{2}\right)^{9}+3\left(\frac{3}{2}\right)^{9}-3 =6\left(\frac{3}{2}\right)^{9}-3.$$

Therefore
$$\alpha=6,\qquad\beta=3,\qquad \alpha+\beta=9.$$
Finally,
$$(\alpha+\beta)^{2}=9^{2}=81.$$

Option C is correct.

To find the sum of all coefficients in a polynomial expansion, substitute $$x = 1$$.

$$A = (1 - 3(1) + 10(1)^2)^n = (1 - 3 + 10)^n = 8^n$$

$$B = (1 + (1)^2)^n = 2^n$$

Now, $$A = 8^n = (2^3)^n = 2^{3n} = (2^n)^3 = B^3$$.

The answer is $$A = B^3$$, which corresponds to Option (1).

Four consecutive coefficients in $$(1+x)^n$$: $$\binom{n}{r-1} = 28-p$$, $$\binom{n}{r} = p$$, $$\binom{n}{r+1} = 70-\alpha$$, $$\binom{n}{r+2} = \alpha$$.

Using the property that consecutive binomial coefficients satisfy: $$\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}$$.

Sum of 1st and 2nd: $$(28-p) + p = 28 = \binom{n+1}{r}$$.

Sum of 3rd and 4th: $$(70-\alpha) + \alpha = 70 = \binom{n+1}{r+2}$$.

Sum of 2nd and 3rd: $$p + (70-\alpha) = \binom{n+1}{r+1}$$.

Now $$\binom{n+1}{r} = 28$$, $$\binom{n+1}{r+2} = 70$$.

Using $$\frac{\binom{n+1}{r+2}}{\binom{n+1}{r}} = \frac{(n+1-r)(n-r)}{(r+1)(r+2)} = \frac{70}{28} = \frac{5}{2}$$.

So $$2(n+1-r)(n-r) = 5(r+1)(r+2)$$.

Try $$n = 7, r = 1$$: $$2(7)(6) = 84$$, $$5(2)(3) = 30$$. No.

Try $$n = 7, r = 2$$: $$2(6)(5) = 60$$, $$5(3)(4) = 60$$.

$$\binom{8}{3} = 56 = p + (70-\alpha)$$, so $$p + 70 - \alpha = 56 \Rightarrow \alpha - p = 14$$.

$$2\alpha - 3p = 70 - 63 = 7$$.

The answer is Option (1): $$\boxed{7}$$.

General term:
$$x^k(1+x)^{100-k},\quad k=2\text{ to }54$$

$$For(x^{70}):pick(x^{70-k}) from ((1+x)^{100-k})$$

Coefficient:
$$\binom{100-k}{70-k}$$

Sum:
$$\sum_{k=2}^{54}\binom{100-k}{70-k}$$

Let (r=70-k):
$$r=68\to16$$
$$\Rightarrow\sum_{r=16}^{68}\binom{30+r}{r}$$

Use identity:
$$\sum_{r=a}^b\binom{n+r}{r}$$
$$=\binom{n+b+1}{b}-\binom{n+a}{a-1}$$

$$=\binom{99}{68}-\binom{46}{15}$$

$$=\binom{99}{p}-\binom{46}{q}$$

$$\Rightarrow p=68,\ q=15p+q=83$$

The coefficient of $$x^k$$ in the expansion of $$(1+x)^n$$ is $${}^{n}C_{k}$$ (the binomial coefficient).

Given that the coefficients of $$x^{4},\,x^{5},\,x^{6}$$ are in arithmetic progression (A.P.), they must satisfy the A.P. condition
$$2\;{}^{n}C_{5}=\,{}^{n}C_{4}+{}^{n}C_{6}\quad -(1)$$

To simplify $$-(1)$$, divide every term by $${}^{n}C_{5}$$:

$$2=\frac{{}^{n}C_{4}}{{}^{n}C_{5}}+\frac{{}^{n}C_{6}}{{}^{n}C_{5}}\quad -(2)$$

Evaluate each ratio separately.
Using $$\displaystyle {}^{n}C_{r}=\frac{n!}{r!\,(n-r)!}$$:

$$\frac{{}^{n}C_{4}}{{}^{n}C_{5}} =\frac{\dfrac{n!}{4!\,(n-4)!}}{\dfrac{n!}{5!\,(n-5)!}} =\frac{5!\,(n-5)!}{4!\,(n-4)!} =5\;\frac{(n-5)!}{(n-4)!} =\frac{5}{\,n-4\,} \quad -(3)$$

$$\frac{{}^{n}C_{6}}{{}^{n}C_{5}} =\frac{\dfrac{n!}{6!\,(n-6)!}}{\dfrac{n!}{5!\,(n-5)!}} =\frac{5!\,(n-5)!}{6!\,(n-6)!} =\frac{(n-5)!}{6\,(n-6)!} =\frac{\,n-5\,}{6}\quad -(4)$$

Substitute $$(3)$$ and $$(4)$$ into $$(2)$$:

$$2=\frac{5}{n-4}+\frac{n-5}{6}$$

Clear denominators by multiplying through by $$6(n-4)$$:

$$12(n-4)=30+(n-5)(n-4)$$

Simplify both sides:

Left side: $$12n-48$$
Right side: $$(n-5)(n-4)=n^{2}-9n+20$$, so
$$30+(n^{2}-9n+20)=n^{2}-9n+50$$

Set the two sides equal:

$$12n-48=n^{2}-9n+50$$

Bring all terms to the right:

$$0=n^{2}-9n+50-12n+48$$
$$\Rightarrow n^{2}-21n+98=0 \quad -(5)$$

Factor or use the quadratic formula on $$(5)$$.
Discriminant: $$\Delta=21^{2}-4\cdot98=441-392=49=7^{2}$$

Hence

$$n=\frac{21\pm7}{2}\;\Longrightarrow\; n_{1}=7,\; n_{2}=14$$

The question asks for the maximum possible integer $$n$$, so

Answer: $$n=14$$ (Option D).

We need the constant term in the expansion of $$\left(\frac{\sqrt[5]{3}}{x} + \frac{2x}{\sqrt[3]{5}}\right)^{12}$$.

Using binomial theorem, the general term is:

$$ T_{r+1} = \binom{12}{r} \left(\frac{3^{1/5}}{x}\right)^{12-r} \left(\frac{2x}{5^{1/3}}\right)^r $$

$$ T_{r+1} = \binom{12}{r} \cdot 3^{(12-r)/5} \cdot x^{-(12-r)} \cdot \frac{2^r \cdot x^r}{5^{r/3}} $$

$$ T_{r+1} = \binom{12}{r} \cdot \frac{3^{(12-r)/5} \cdot 2^r}{5^{r/3}} \cdot x^{r-(12-r)} = \binom{12}{r} \cdot \frac{3^{(12-r)/5} \cdot 2^r}{5^{r/3}} \cdot x^{2r-12} $$

For the constant term, set $$2r - 12 = 0 \Rightarrow r = 6$$. Thus

$$ T_7 = \binom{12}{6} \cdot \frac{3^{(12-6)/5} \cdot 2^6}{5^{6/3}} $$

which becomes

$$ T_7 = 924 \cdot \frac{3^{6/5} \cdot 64}{5^2} $$

and simplifies to

$$ T_7 = 924 \cdot \frac{64 \cdot 3^{6/5}}{25}. $$

Since $$3^{6/5} = 3^{1+1/5} = 3 \cdot 3^{1/5} = 3 \cdot \sqrt[5]{3},$$ we get

$$ T_7 = 924 \cdot \frac{64 \cdot 3 \cdot \sqrt[5]{3}}{25} = 924 \cdot \frac{192 \cdot \sqrt[5]{3}}{25} = \frac{924 \times 192}{25} \cdot \sqrt[5]{3}. $$

Computing $$924 \times 192 = 924 \times 200 - 924 \times 8 = 184800 - 7392 = 177408$$ gives

$$ T_7 = \frac{177408}{25} \cdot \sqrt[5]{3}. $$

This equals $$\alpha \times 2^8 \times \sqrt[5]{3} = 256\alpha \cdot \sqrt[5]{3},$$ so

$$256\alpha = \frac{177408}{25}$$

and

$$\alpha = \frac{177408}{25 \times 256} = \frac{177408}{6400} = 27.72.$$

Multiplying by 25 yields

$$25\alpha = \frac{177408}{256} = 693.$$

Verifying,

$$693 \times 256 = 693 \times 250 + 693 \times 6 = 173250 + 4158 = 177408.$$

Correct!

The correct answer is Option (4): 693.

We need to find the term independent of $$x$$ in the expansion of $$\left(\sqrt{a}x^2 + \frac{1}{2x^3}\right)^{10}$$.

The general term using the Binomial Theorem is:

$$T_{r+1} = \binom{10}{r}(\sqrt{a}x^2)^{10-r}\left(\frac{1}{2x^3}\right)^r$$

$$= \binom{10}{r}(\sqrt{a})^{10-r} \cdot \frac{1}{2^r} \cdot x^{2(10-r)-3r}$$

$$= \binom{10}{r}\frac{a^{(10-r)/2}}{2^r} \cdot x^{20-5r}$$

For the term independent of $$x$$: $$20 - 5r = 0 \Rightarrow r = 4$$.

The term independent of $$x$$ is:

$$T_5 = \binom{10}{4}\frac{a^{3}}{2^4} = 210 \cdot \frac{a^3}{16} = \frac{210a^3}{16} = \frac{105a^3}{8}$$

Setting this equal to 105:

$$\frac{105a^3}{8} = 105$$

$$a^3 = 8$$

$$a = 2$$

Therefore $$a^2 = 4$$.

The correct answer is Option 2: 4.

The binomial expansion is given by: $$\left(\frac{1}{3}x^{1/3} + \frac{1}{2x^{2/3}}\right)^{18}$$.

The general term in the expansion of $$(a + b)^k$$ is $$T_{r+1} = \binom{k}{r} a^{k-r} b^r$$.

Here, $$a = \frac{1}{3}x^{1/3}$$, $$b = \frac{1}{2} x^{-2/3}$$, and $$k = 18$$.

So, the general term is: $$T_{r+1} = \binom{18}{r} \left( \frac{1}{3} x^{1/3} \right)^{18-r} \left( \frac{1}{2} x^{-2/3} \right)^r$$

Simplifying: $$T_{r+1} = \binom{18}{r} \left( \frac{1}{3} \right)^{18-r} (x^{1/3})^{18-r} \cdot \left( \frac{1}{2} \right)^r (x^{-2/3})^r = \binom{18}{r} \left( \frac{1}{3} \right)^{18-r} \left( \frac{1}{2} \right)^r x^{\frac{18-r}{3} - \frac{2r}{3}}$$

The exponent of $$x$$ is: $$\frac{18-r}{3} - \frac{2r}{3} = \frac{18 - r - 2r}{3} = \frac{18 - 3r}{3} = 6 - r$$

Thus: $$T_{r+1} = \binom{18}{r} \left( \frac{1}{3} \right)^{18-r} \left( \frac{1}{2} \right)^r x^{6 - r}$$

The coefficient of $$T_{r+1}$$ is: $$\binom{18}{r} \left( \frac{1}{3} \right)^{18-r} \left( \frac{1}{2} \right)^r$$

The seventh term corresponds to $$r+1 = 7$$, so $$r = 6$$.

The thirteenth term corresponds to $$r+1 = 13$$, so $$r = 12$$.

Let $$m$$ be the coefficient of the seventh term and $$n$$ be the coefficient of the thirteenth term.

So: $$m = \binom{18}{6} \left( \frac{1}{3} \right)^{12} \left( \frac{1}{2} \right)^6$$ $$n = \binom{18}{12} \left( \frac{1}{3} \right)^{6} \left( \frac{1}{2} \right)^{12}$$

Since $$\binom{18}{12} = \binom{18}{6}$$, we have: $$n = \binom{18}{6} \left( \frac{1}{3} \right)^{6} \left( \frac{1}{2} \right)^{12}$$

Now, compute $$\frac{n}{m}$$: $$\frac{n}{m} = \frac{ \binom{18}{6} \left( \frac{1}{3} \right)^{6} \left( \frac{1}{2} \right)^{12} }{ \binom{18}{6} \left( \frac{1}{3} \right)^{12} \left( \frac{1}{2} \right)^{6} } = \frac{ \left( \frac{1}{3} \right)^{6} \left( \frac{1}{2} \right)^{12} }{ \left( \frac{1}{3} \right)^{12} \left( \frac{1}{2} \right)^{6} } = \left( \frac{1}{3} \right)^{6-12} \left( \frac{1}{2} \right)^{12-6} = \left( \frac{1}{3} \right)^{-6} \left( \frac{1}{2} \right)^{6} = 3^{6} \cdot 2^{-6} = \left( \frac{3}{2} \right)^6$$

Now, compute $$\left( \frac{n}{m} \right)^{1/3}$$: $$\left( \frac{n}{m} \right)^{1/3} = \left( \left( \frac{3}{2} \right)^6 \right)^{1/3} = \left( \frac{3}{2} \right)^{6 \cdot \frac{1}{3}} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$$

The value is $$\frac{9}{4}$$, which corresponds to option D.

$$\left(x^{2/3}+\frac{1}{2}x^{-2/5}\right)^9$$. General term: $$T_{r+1} = \binom{9}{r}(x^{2/3})^{9-r}\left(\frac{1}{2}x^{-2/5}\right)^r = \binom{9}{r}\frac{1}{2^r}x^{2(9-r)/3-2r/5}$$.

Power of x: $$\frac{2(9-r)}{3} - \frac{2r}{5} = \frac{10(9-r)-6r}{15} = \frac{90-16r}{15}$$.

For $$x^{2/3}$$: $$\frac{90-16r}{15} = \frac{2}{3} = \frac{10}{15}$$. $$90-16r = 10$$. $$r = 5$$.

Coefficient: $$\binom{9}{5}/2^5 = 126/32 = 63/16$$.

For $$x^{-2/5}$$: $$\frac{90-16r}{15} = -\frac{2}{5} = -\frac{6}{15}$$. $$90-16r = -6$$. $$r = 6$$.

Coefficient: $$\binom{9}{6}/2^6 = 84/64 = 21/16$$.

Sum = $$63/16 + 21/16 = 84/16 = 21/4$$.

The correct answer is Option 1: 21/4.

General term: T(r+1) = C(15,r) × 2^((15-r)/5) × 5^(r/3)

For rational terms: (15-r)/5 and r/3 must be non-negative integers.

15-r must be divisible by 5: r = 0, 5, 10, 15

r must be divisible by 3: r = 0, 3, 6, 9, 12, 15

Common values: r = 0 and r = 15.

T(1) = C(15,0) × 2³ × 1 = 8

T(16) = C(15,15) × 1 × 5⁵ = 3125

Sum = 8 + 3125 = 3133

The correct answer is Option 1: 3133.

 $$a = (x+3)^{n-1}$$

$$ r = \frac{x+2}{x+3} $$

$$ S = a \cdot \frac{1 - r^n}{1 - r} $$

$$ S = (x+3)^{n-1} \cdot \frac{1 - \left(\frac{x+2}{x+3}\right)^n}{1 - \frac{x+2}{x+3}} $$

$$ 1 - \frac{x+2}{x+3} = \frac{(x+3) - (x+2)}{x+3} = \frac{1}{x+3} $$

$$ 1 - \left(\frac{x+2}{x+3}\right)^n = \frac{(x+3)^n - (x+2)^n}{(x+3)^n} $$

$$ S = (x+3)^{n-1} \cdot \frac{(x+3)^n - (x+2)^n}{(x+3)^n} \cdot (x+3) = (x+3)^n - (x+2)^n $$

To find the sum of all coefficients of a polynomial $$f(x)$$, we substitute $$x = 1$$

$$ \sum_{r=0}^{n} \alpha_r = S(1) = (1+3)^n - (1+2)^n = 4^n - 3^n $$

This is because when we substitute $$x = 1$$, every term $$\alpha_r x^r$$ becomes just $$\alpha_r$$, so their sum equals the polynomial evaluated at $$x = 1$$.

Finally, comparing $$4^n - 3^n = \beta^n - \gamma^n$$ with $$\beta, \gamma \in \mathbb{N}$$, we identify

$$\beta = 4$$ and $$\gamma = 3$$.

$$ \beta^2 + \gamma^2 = 16 + 9 = 25 $$

$$(1-x)^{2008}(1+x+x^2)^{2007}$$

Note: $$1 + x + x^2 = \frac{1-x^3}{1-x}$$ (for $$x \neq 1$$).

$$(1-x)^{2008} \cdot \left(\frac{1-x^3}{1-x}\right)^{2007} = (1-x)^{2008-2007} \cdot (1-x^3)^{2007} = (1-x)(1-x^3)^{2007}$$

We need the coefficient of $$x^{2012}$$ in $$(1-x)(1-x^3)^{2007}$$.

$$= \text{coeff of } x^{2012} \text{ in } (1-x^3)^{2007} - \text{coeff of } x^{2011} \text{ in } (1-x^3)^{2007}$$

Coeff of $$x^{3k}$$ in $$(1-x^3)^{2007}$$ is $$(-1)^k\binom{2007}{k}$$.

$$x^{2012}$$: $$2012/3$$ is not an integer, so coefficient = 0.

$$x^{2011}$$: $$2011/3$$ is not an integer, so coefficient = 0.

Answer: $$0 - 0 = 0$$.

The answer is $$\boxed{0}$$.

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.

Find the constant term in $$(1 + 2x - 3x^3)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$$.

General term of $$\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$$:

$$ T_{r+1} = \binom{9}{r}\left(\frac{3}{2}x^2\right)^{9-r}\left(-\frac{1}{3x}\right)^r = \binom{9}{r}\frac{3^{9-r}}{2^{9-r}} \cdot \frac{(-1)^r}{3^r} \cdot x^{2(9-r)-r} $$

$$ = \binom{9}{r}\frac{(-1)^r \cdot 3^{9-2r}}{2^{9-r}} x^{18-3r} $$

The constant term in the full product comes from:

1. $$1 \times$$ (term with $$x^0$$ from the expansion): $$18 - 3r = 0 \Rightarrow r = 6$$.

2. $$2x \times$$ (term with $$x^{-1}$$): $$18 - 3r = -1 \Rightarrow 3r = 19$$. Not integer. No contribution.

3. $$-3x^3 \times$$ (term with $$x^{-3}$$): $$18 - 3r = -3 \Rightarrow r = 7$$.

Contribution 1 ($$r = 6$$):

$$ 1 \times \binom{9}{6}\frac{(-1)^6 \cdot 3^{-3}}{2^3} = 84 \times \frac{1}{27 \times 8} = 84 \times \frac{1}{216} = \frac{84}{216} = \frac{7}{18} $$

Contribution 3 ($$r = 7$$):

$$ -3 \times \binom{9}{7}\frac{(-1)^7 \cdot 3^{-5}}{2^2} = -3 \times 36 \times \frac{-1}{243 \times 4} = -3 \times 36 \times \frac{-1}{972} = \frac{108}{972} = \frac{1}{9} $$

Total constant term:

$$ p = \frac{7}{18} + \frac{1}{9} = \frac{7}{18} + \frac{2}{18} = \frac{9}{18} = \frac{1}{2} $$

$$ 108p = 108 \times \frac{1}{2} = 54 $$

The answer is 54.

$$(x+y)^n$$: $$T_2 = nx^{n-1}y = 135$$, $$T_3 = \binom{n}{2}x^{n-2}y^2 = 30$$, $$T_4 = \binom{n}{3}x^{n-3}y^3 = 10/3$$.

$$T_3/T_2 = \frac{(n-1)y}{2x} = \frac{30}{135} = \frac{2}{9}$$.

$$T_4/T_3 = \frac{(n-2)y}{3x} = \frac{10/3}{30} = \frac{1}{9}$$.

Dividing: $$\frac{2(n-2)}{3(n-1)} = \frac{1/9}{2/9} = 1/2$$. So $$4(n-2) = 3(n-1) \Rightarrow n = 5$$.

$$y/x = \frac{2\times2}{9\times4} = 1/9$$... From $$\frac{4y}{2x} = 2/9$$: $$y/x = 1/9$$.

From $$T_2 = 5x^4y = 135$$: $$5x^4(x/9) = 135 \Rightarrow 5x^5/9 = 135 \Rightarrow x^5 = 243 = 3^5 \Rightarrow x = 3$$. $$y = 1/3$$.

$$6(n^3+x^2+y) = 6(125+9+1/3) = 6(134+1/3) = 6(403/3) = 806$$.

The answer is 806.

We need to find the sum of the coefficients of $$x^3$$ and $$x^{-13}$$ in $$(1+x)(1-x)^2\left(1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}\right)^5$$.

Recognizing the binomial pattern, we note that $$1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3} = \left(1 + \frac{1}{x}\right)^3 = \frac{(x+1)^3}{x^3}$$, so that $$\left(1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}\right)^5 = \frac{(x+1)^{15}}{x^{15}}$$.

Thus the product becomes $$(1+x)(1-x)^2 \cdot \frac{(x+1)^{15}}{x^{15}} = \frac{(1+x)^{16}(1-x)^2}{x^{15}}$$. Let $$P(x) = (1+x)^{16}(1-x)^2$$; then the coefficient of $$x^k$$ in the original expression is the coefficient of $$x^{k+15}$$ in $$P(x)$$.

To find the coefficient of $$x^3$$, we look for $$[x^{18}]$$ in $$P(x) = (1+x)^{16}(1 - 2x + x^2)$$. Since $$(1+x)^{16}$$ has degree 16, the only way to obtain $$x^{18}$$ is via $$\binom{16}{16}x^{16} \times x^2 = x^{18}$$. Hence $$[x^{18}]P(x) = \binom{16}{16} \times 1 = 1$$.

For the coefficient of $$x^{-13}$$ we need $$[x^2]$$ in $$P(x)$$, which is given by $$[x^2]P(x) = \binom{16}{2}(1) + \binom{16}{1}(-2) + \binom{16}{0}(1) = 120 - 32 + 1 = 89$$.

Adding these results yields $$1 + 89 = \mathbf{90}$$.

The answer is $$\mathbf{90}$$.

Consider the expansion of $$(7^{1/2} + 11^{1/6})^{824}$$. The general term of this expansion is given by $$ T_{r+1} = \binom{824}{r} (7^{1/2})^{824-r} (11^{1/6})^r = \binom{824}{r} 7^{(824-r)/2} \cdot 11^{r/6} $$.

For such a term to be integral, both exponents must be integers. The condition $$\frac{824 - r}{2} \in \mathbb{Z}$$ implies that $$r$$ must be even, while $$\frac{r}{6} \in \mathbb{Z}$$ implies that $$r$$ must be divisible by 6.

Hence $$r$$ must be a multiple of 6 subject to $$0 \leq r \leq 824$$, namely $$r = 0, 6, 12, \ldots, 822$$.

The total number of integral terms is given by $$\frac{822 - 0}{6} + 1 = 137 + 1 = 138$$.

Therefore, the answer is $$\boxed{138}$$.

We start with the two sums
$$\alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^{2}}{k+1},\qquad \beta = \sum_{k=0}^{\,n-1} \frac{\binom{n}{k}\,\binom{n}{k+1}}{k+2}.$$

Case 1: Closed form for $$\alpha$$
First use the relation
$$\frac{\binom{n}{k}}{k+1}=\frac{1}{\,n+1\,}\binom{\,n+1\,}{\,k+1\,}.$$ Hence

$$ \alpha =\frac{1}{n+1}\sum_{k=0}^{n}\binom{n}{k}\binom{\,n+1\,}{\,k+1\,}. $$

Notice that $$\binom{\,n+1\,}{\,k+1\,}=\binom{\,n+1\,}{\,n-k\,}.$$
Putting $$j=n-k$$ the summation becomes $$ \alpha=\frac{1}{n+1}\sum_{j=0}^{n}\binom{n}{j}\binom{\,n+1\,}{\,j\,}. $$

By Vandermonde’s identity
$$ \sum_{j=0}^{n}\binom{n}{j}\binom{\,n+1\,}{\,j\,} =\binom{\,2n+1\,}{\,n\,}. $$ Therefore

$$ \boxed{\alpha=\dfrac{1}{\,n+1\,}\binom{\,2n+1\,}{\,n\,}.} $$

Case 2: Closed form for $$\beta$$

Rewrite the general term of $$\beta$$ with a shifted index $$r=k+1$$ ($$1\le r\le n$$):
$$ \beta=\sum_{r=1}^{n}\frac{\binom{n}{r-1}\,\binom{n}{r}}{r+1}. $$

Use $$ (r+1)\binom{n}{r}=n\binom{\,n-1\,}{\,r-1\,}. $$ Multiplying numerator and denominator of each term by $$r+1$$ gives

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

Put $$s=r-1$$ (so $$0\le s\le n-1$$):
$$ \beta=\frac{n}{(n+1)(n+2)} \sum_{s=0}^{n-1}\binom{\,n-1\,}{\,s\,}\binom{\,n+2\,}{\,s+2\,}. $$

Since $$\binom{\,n+2\,}{\,s+2\,}=\binom{\,n+2\,}{\,n-s\,},$$ the sum equals the coefficient of $$x^{\,n}$$ in $$(1+x)^{\,n-1}(1+x)^{\,n+2}=(1+x)^{\,2n+1},$$ which by the binomial theorem is $$\binom{\,2n+1\,}{\,n\,}.$$ Hence

$$ \boxed{\beta=\dfrac{n}{(n+1)(n+2)}\binom{\,2n+1\,}{\,n\,}.} $$

Case 3: Using the given relation
The problem states $$5\alpha = 6\beta.$$ Insert the closed forms:

$$ 5\left[\frac{1}{n+1}\binom{\,2n+1\,}{\,n\,}\right] =6\left[\frac{n}{(n+1)(n+2)}\binom{\,2n+1\,}{\,n\,}\right]. $$

Because $$\binom{\,2n+1\,}{\,n\,}\neq 0$$, it cancels out, and so does the common factor $$n+1$$, giving

$$ 5 = 6\,\frac{n}{n+2}. $$

Solve for $$n$$:
$$ 5(n+2)=6n \;\;\Longrightarrow\;\; 5n+10=6n \;\;\Longrightarrow\;\; n=10. $$

Hence $$n=10$$.

Write the first expression in binomial form:
$$\left(ax^{2} + \frac{70}{27\,b\,x}\right)^{4}$$

General term of $$\left(u+v\right)^{4}$$ is $$\binom{4}{k}u^{k}v^{4-k}$$.
Take $$u = ax^{2}$$ and $$v = \dfrac{70}{27\,b\,x}$$.

The term containing $$k$$ copies of $$u$$ and $$4-k$$ copies of $$v$$ equals
$$\binom{4}{k}\,a^{k}\left(ax^{2}\right)^{k} \left(\frac{70}{27\,b\,x}\right)^{4-k} = \binom{4}{k}\,a^{k} \left(\frac{70}{27\,b}\right)^{4-k} x^{\,2k-(4-k)} = \binom{4}{k}\,a^{k} \left(\frac{70}{27\,b}\right)^{4-k} x^{\,3k-4}$$

We need the power of $$x$$ equal to $$5$$:
$$3k-4 = 5 \;\Longrightarrow\; 3k = 9 \;\Longrightarrow\; k = 3$$

Substituting $$k=3$$ gives the desired coefficient (call it $$C_1$$):
$$C_1 = \binom{4}{3}a^{3}\left(\frac{70}{27\,b}\right)^{1} = 4 \cdot a^{3}\cdot\frac{70}{27\,b} = \frac{280\,a^{3}}{27\,b}$$

Now treat the second expression:
$$\left(ax - \frac{1}{b\,x^{2}}\right)^{7}$$

Here $$u = ax$$, $$v = -\dfrac{1}{b\,x^{2}}$$ and the general term is
$$\binom{7}{r}u^{r}v^{7-r} = \binom{7}{r}a^{r}(ax)^{\,r} \left(-\frac{1}{b\,x^{2}}\right)^{7-r} = \binom{7}{r}a^{r}(-1)^{7-r}b^{-(7-r)} x^{\,r-2(7-r)} = \binom{7}{r}a^{r}(-1)^{7-r}b^{-(7-r)} x^{\,3r-14}$$

We need the power of $$x$$ equal to $$-5$$:
$$3r-14 = -5 \;\Longrightarrow\; 3r = 9 \;\Longrightarrow\; r = 3$$

With $$r=3$$, the power of $$-1$$ is $$(-1)^{7-3}=(-1)^{4}=+1$$.
The required coefficient (call it $$C_2$$) is
$$C_2 = \binom{7}{3}a^{3}b^{-(7-3)} = 35\,a^{3}\,b^{-4} = \frac{35\,a^{3}}{b^{4}}$$

According to the question, these two coefficients are equal:

$$\frac{280\,a^{3}}{27\,b} = \frac{35\,a^{3}}{b^{4}}$$

Cancel the common non-zero factor $$a^{3}$$ and cross-multiply:

$$280\,b^{3} = 35 \times 27$$

Divide by $$35$$:

$$b^{3} = \frac{27}{8}$$

Taking the real cube root (since $$b$$ is real and $$\dfrac{27}{8} \gt 0$$):

$$b = \frac{3}{2}$$

Finally, the required value is $$2b$$:

$$2b = 2 \times \frac{3}{2} = 3$$

Therefore, the answer is 3.

The coefficient of $$x^{10-r}$$ in $$(1+x)^{10}$$ is $$a_r = \binom{10}{10-r}$$.

Similarly, $$a_{r-1} = \binom{10}{11-r}$$.

$$\frac{a_r}{a_{r-1}} = \frac{\binom{10}{10-r}}{\binom{10}{11-r}} = \frac{\frac{10!}{(10-r)!r!}}{\frac{10!}{(11-r)!(r-1)!}} = \frac{(11-r)!(r-1)!}{(10-r)!r!} = \frac{11-r}{r}$$

So $$\left(\frac{a_r}{a_{r-1}}\right)^2 = \frac{(11-r)^2}{r^2}$$

$$\sum_{r=1}^{10} r^3 \cdot \frac{(11-r)^2}{r^2} = \sum_{r=1}^{10} r(11-r)^2$$

$$= \sum_{r=1}^{10} r(121 - 22r + r^2) = 121\sum r - 22\sum r^2 + \sum r^3$$

Using standard formulas: $$\sum_{r=1}^{10} r = 55$$, $$\sum_{r=1}^{10} r^2 = 385$$, $$\sum_{r=1}^{10} r^3 = 3025$$.

$$= 121(55) - 22(385) + 3025 = 6655 - 8470 + 3025 = 1210$$

The correct answer is Option 2: 1210.

Step 1: Find the coefficient of $$x^{15}$$ in the first expansion

The first binomial expression is:

$$\left(ax^3 + \frac{1}{bx^{\frac{1}{3}}}\right)^{15}$$

The general term $$T_{r+1}$$ in this expansion is given by:

$$T_{r+1} = \binom{15}{r} (ax^3)^{15-r} \left(\frac{1}{bx^{\frac{1}{3}}}\right)^r$$

$$T_{r+1} = \binom{15}{r} a^{15-r} b^{-r} x^{3(15-r) - \frac{r}{3}}$$

$$T_{r+1} = \binom{15}{r} a^{15-r} b^{-r} x^{45 - 3r - \frac{r}{3}}$$

$$T_{r+1} = \binom{15}{r} a^{15-r} b^{-r} x^{45 - \frac{10r}{3}}$$

To find the coefficient of $$x^{15}$$, we set the exponent of $$x$$ equal to 15:

$$45 - \frac{10r}{3} = 15$$

$$\frac{10r}{3} = 30$$

$$10r = 90$$

$$r = 9$$

Substituting $$r = 9$$ back into the term gives the coefficient of $$x^{15}$$:

$$\text{Coefficient of } x^{15} = \binom{15}{9} a^{15-9} b^{-9} = \binom{15}{9} \frac{a^6}{b^9}$$

Step 2: Find the coefficient of $$x^{-15}$$ in the second expansion

The second binomial expression is:

$$\left(ax^{\frac{1}{3}} - \frac{1}{bx^3}\right)^{15}$$

The general term $$T_{k+1}$$ in this expansion is given by:

$$T_{k+1} = \binom{15}{k} \left(ax^{\frac{1}{3}}\right)^{15-k} \left(-\frac{1}{bx^3}\right)^k$$

$$T_{k+1} = \binom{15}{k} a^{15-k} (-1)^k b^{-k} x^{\frac{15-k}{3} - 3k}$$

$$T_{k+1} = \binom{15}{k} a^{15-k} (-1)^k b^{-k} x^{5 - \frac{k}{3} - 3k}$$

$$T_{k+1} = \binom{15}{k} a^{15-k} (-1)^k b^{-k} x^{5 - \frac{10k}{3}}$$

To find the coefficient of $$x^{-15}$$, we set the exponent of $$x$$ equal to -15:

$$5 - \frac{10k}{3} = -15$$

$$20 = \frac{10k}{3}$$

$$60 = 10k$$

$$k = 6$$

Substituting $$k = 6$$ back into the term gives the coefficient of $$x^{-15}$$:

$$\text{Coefficient of } x^{-15} = \binom{15}{6} a^{15-6} (-1)^6 b^{-6} = \binom{15}{6} \frac{a^9}{b^6}$$

Step 3: Equate the two coefficients and solve for the relationship

We are given that the two coefficients are equal:

$$\binom{15}{9} \frac{a^6}{b^9} = \binom{15}{6} \frac{a^9}{b^6}$$

Using the identity $$\binom{n}{r} = \binom{n}{n-r}$$, we know that $$\binom{15}{9} = \binom{15}{15-9} = \binom{15}{6}$$. Therefore, these terms cancel out from both sides:

$$\frac{a^6}{b^9} = \frac{a^9}{b^6}$$

Cross-multiplying the terms gives:

$$a^6 b^6 = a^9 b^9$$

$$1 = a^3 b^3$$

$$(ab)^3 = 1$$

Since $$a$$ and $$b$$ are positive real numbers, taking the cube root of both sides gives:

$$ab = 1$$

Conclusion:

the correct option is $$ab = 1$$.

In $$\left(ax - \frac{1}{bx^2}\right)^{13}$$, the general term is:

$$T_{r+1} = \binom{13}{r}(ax)^{13-r}\left(\frac{-1}{bx^2}\right)^r = \binom{13}{r}a^{13-r}\frac{(-1)^r}{b^r}x^{13-3r}$$

For $$x^7$$: $$13 - 3r = 7 \Rightarrow r = 2$$

Coefficient = $$\binom{13}{2}a^{11}\frac{1}{b^2} = 78\frac{a^{11}}{b^2}$$

In $$\left(ax + \frac{1}{bx^2}\right)^{13}$$, the general term is:

$$T_{r+1} = \binom{13}{r}(ax)^{13-r}\left(\frac{1}{bx^2}\right)^r = \binom{13}{r}\frac{a^{13-r}}{b^r}x^{13-3r}$$

For $$x^{-5}$$: $$13 - 3r = -5 \Rightarrow r = 6$$

Coefficient = $$\binom{13}{6}\frac{a^7}{b^6} = 1716\frac{a^7}{b^6}$$

Setting coefficients equal:

$$78\frac{a^{11}}{b^2} = 1716\frac{a^7}{b^6}$$

$$78 a^4 b^4 = 1716$$

$$a^4 b^4 = \frac{1716}{78} = 22$$

The correct answer is Option 3: 22.

We need to find the ordered pair $$(l, n)$$ given the expression $$\dfrac{{}^{200}C_{99} \cdot K}{a} = \dfrac{2^l \cdot m}{n}$$ where $$m$$ and $$n$$ are odd.

Put $$x = 1$$: $$(1+1)^{99} = 2^{99}$$ (sum of all coefficients).

Put $$x = -1$$: $$(1-1)^{99} = 0$$ (alternating sum).

Sum of coefficients of odd powers = $$\dfrac{2^{99} - 0}{2} = 2^{98}$$.

So $$K = 2^{98}$$.

The expansion has 201 terms, so the middle term is the 101st term ($$r = 100$$):

$$a = {}^{200}C_{100} \cdot 2^{100} \cdot \left(\frac{1}{\sqrt{2}}\right)^{100} = {}^{200}C_{100} \cdot 2^{100} \cdot 2^{-50} = {}^{200}C_{100} \cdot 2^{50}$$

$$\frac{{}^{200}C_{99} \cdot K}{a} = \frac{{}^{200}C_{99} \cdot 2^{98}}{{}^{200}C_{100} \cdot 2^{50}}$$

Now, $$\dfrac{{}^{200}C_{99}}{{}^{200}C_{100}} = \dfrac{100!\ \cdot\ 100!}{99!\ \cdot\ 101!} = \dfrac{100}{101}$$.

$$= \frac{100}{101} \cdot 2^{48}$$

$$100 = 2^2 \times 25$$, so:

$$\frac{100}{101} \cdot 2^{48} = \frac{25 \cdot 2^2 \cdot 2^{48}}{101} = \frac{2^{50} \cdot 25}{101}$$

Here $$m = 25$$ (odd) and $$n = 101$$ (odd), and $$l = 50$$.

The ordered pair $$(l, n) = (50, 101)$$.

The correct answer is Option C: $$(50, 101)$$.

Use the identity:
$$\sum_{r=0}^n\binom{n}{r}\binom{m}{r}=\binom{n+m}{n}$$

Here (n=22), (m=23). So,
$$\sum_{r=0}^{22}\binom{22}{r}\binom{23}{r}=\binom{45}{22}$$

$$\binom{45}{22}$$

We need to find the relation between $$a$$ and $$b$$ given that the coefficient of $$x^7$$ in $$\left(ax^2 + \dfrac{1}{2bx}\right)^{11}$$ equals the coefficient of $$x^{-7}$$ in $$\left(ax - \dfrac{1}{3bx^2}\right)^{11}$$.

To determine the coefficient of $$x^7$$ in the expansion of $$\left(ax^2 + \dfrac{1}{2bx}\right)^{11}$$, consider the general term $$T_{r+1} = \binom{11}{r}(ax^2)^{11-r}\left(\dfrac{1}{2bx}\right)^r = \binom{11}{r}\dfrac{a^{11-r}}{(2b)^r} \cdot x^{22-3r}\,.$$ Requiring the exponent to be 7 gives $$22 - 3r = 7 \Rightarrow r = 5$$, so the coefficient is $$\binom{11}{5}\dfrac{a^6}{(2b)^5} = \dfrac{462 \cdot a^6}{32b^5}\,.$$

Similarly, for the coefficient of $$x^{-7}$$ in the expansion of $$\left(ax - \dfrac{1}{3bx^2}\right)^{11}$$, the general term is $$T_{r+1} = \binom{11}{r}(ax)^{11-r}\left(\dfrac{-1}{3bx^2}\right)^r = \binom{11}{r}\dfrac{a^{11-r}(-1)^r}{(3b)^r} \cdot x^{11-3r}\,. $$ Setting $$11 - 3r = -7$$ gives $$r = 6$$ and the coefficient becomes $$\binom{11}{6}\dfrac{a^5 \cdot (-1)^6}{(3b)^6} = \dfrac{462 \cdot a^5}{729b^6}\,.$$

Equating these two coefficients yields

$$\frac{462 \cdot a^6}{32b^5} = \frac{462 \cdot a^5}{729b^6}$$

Cancelling $$462 \cdot a^5 / b^5$$ from both sides leads to

$$\frac{a}{32} = \frac{1}{729b}$$

$$729ab = 32$$

Therefore, the required relation is $$729ab = 32$$, which corresponds to Option A.

Let $$a = \sqrt[4]{2} = 2^{1/4}$$ and $$b = \frac{1}{\sqrt[4]{3}} = 3^{-1/4}$$.

The fifth term from the beginning is: $$T_5 = \binom{n}{4} a^{n-4} b^4$$

The fifth term from the end is: $$T_{n-3} = \binom{n}{4} a^4 b^{n-4}$$

The ratio is:

$$ \frac{T_5}{T_{n-3}} = \frac{a^{n-4} b^4}{a^4 b^{n-4}} = \left(\frac{a}{b}\right)^{n-8} = \left(2^{1/4} \cdot 3^{1/4}\right)^{n-8} = 6^{(n-8)/4} $$

Given this ratio equals $$\sqrt{6} = 6^{1/2}$$:

$$ \frac{n-8}{4} = \frac{1}{2} \implies n - 8 = 2 \implies n = 10 $$

The third term from the beginning is:

$$ T_3 = \binom{10}{2} \cdot (2^{1/4})^8 \cdot (3^{-1/4})^2 = 45 \cdot 2^2 \cdot 3^{-1/2} = 45 \cdot \frac{4}{\sqrt{3}} = \frac{180}{\sqrt{3}} = 60\sqrt{3} $$

The correct answer is $$60\sqrt{3}$$.

We need to evaluate $$\sum_{k=0}^{6} \binom{51-k}{3}$$.

Expanding the sum:

$$ \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3} + \binom{46}{3} + \binom{45}{3} $$

We use the Hockey Stick Identity (or Christmas Stocking Identity):

$$ \sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1} $$

Rearranging our sum in ascending order:

$$ \binom{45}{3} + \binom{46}{3} + \binom{47}{3} + \binom{48}{3} + \binom{49}{3} + \binom{50}{3} + \binom{51}{3} $$

By the Hockey Stick Identity with $$r = 3$$:

$$ \sum_{i=3}^{51} \binom{i}{3} = \binom{52}{4} $$

Our sum starts from $$i = 45$$, not $$i = 3$$. So:

$$ \sum_{i=45}^{51} \binom{i}{3} = \sum_{i=3}^{51} \binom{i}{3} - \sum_{i=3}^{44} \binom{i}{3} = \binom{52}{4} - \binom{45}{4} $$

Therefore: $$\sum_{k=0}^{6} \binom{51-k}{3} = \binom{52}{4} - \binom{45}{4}$$

This matches Option C: $${}^{52}C_4 - {}^{45}C_4$$.

Each subsequent term is formed by multiplying the previous term by

$$\frac{x}{1+x}$$

Hence, the given series is a geometric progression (G.P.).

First term:

$$a = (1+x)^{500}$$

Common ratio:

$$r = \frac{x}{1+x}$$

Number of terms:

$$n = 501$$

The sum of a finite G.P. is

$$S_n = a\left(\frac{1-r^n}{1-r}\right)$$

Substituting the values:

$$S = (1+x)^{500}\left(\frac{1-\left(\frac{x}{1+x}\right)^{501}}{1-\frac{x}{1+x}}\right)$$

Simplifying the denominator:

$$1-\frac{x}{1+x} = \frac{1+x-x}{1+x} = \frac{1}{1+x}$$

Therefore,

$$S = \frac{(1+x)^{500}\left[1-\frac{x^{501}}{(1+x)^{501}}\right]}{\frac{1}{1+x}}$$

Dividing by a fraction means multiplying by its reciprocal:

$$S = (1+x)^{500}(1+x)\left[1-\frac{x^{501}}{(1+x)^{501}}\right]$$

$$S = (1+x)^{501}\left[1-\frac{x^{501}}{(1+x)^{501}}\right]$$

Distributing:

$$S = (1+x)^{501} - (1+x)^{501}\cdot\frac{x^{501}}{(1+x)^{501}}$$

$$S = (1+x)^{501} - x^{501}$$

We need the coefficient of $$x^{301}$$.

The term $$-x^{501}$$ does not affect the coefficient of $$x^{301}$$.

Hence, we only need the coefficient of $$x^{301}$$ in

$$(1+x)^{501}$$

Using the Binomial Theorem:

$$(1+x)^n = \sum_{k=0}^{n}\binom{n}{k}x^k$$

Therefore, the coefficient of $$x^{301}$$ is

$$\binom{501}{301}$$

Using symmetry of binomial coefficients,

$$\binom{501}{301} = \binom{501}{200}$$

Hence, the required option is D.

$$S = 1 \cdot \binom{30}{1}^2 + 2 \cdot \binom{30}{2}^2 + 3 \cdot \binom{30}{3}^2 + \dots + 30 \cdot \binom{30}{30}^2$$

$$S = \sum_{r=1}^{30} r \cdot \binom{30}{r}^2$$

Using the property $$\binom{30}{r} = \binom{30}{30-r}$$:

$$S = \sum_{r=1}^{30} \left[ r \cdot \binom{30}{r} \cdot \binom{30}{30-r} \right]$$

Now substitute $$r \cdot \binom{30}{r} = 30 \cdot \binom{29}{r-1}$$:

$$S = \sum_{r=1}^{30} \left[ 30 \cdot \binom{29}{r-1} \cdot \binom{30}{30-r} \right]$$

$$S = 30 \sum_{r=1}^{30} \binom{29}{r-1} \cdot \binom{30}{30-r}$$

$$\sum_{r=1}^{30} \binom{29}{r-1} \cdot \binom{30}{30-r} = \binom{29+30}{29} = \binom{59}{29}$$

$$S = 30 \cdot \binom{59}{29}$$

$$S = 30 \cdot \frac{59!}{29! \cdot (59-29)!} = 30 \cdot \frac{59!}{29! \cdot 30!}$$

$$S = \frac{1}{2} \cdot \frac{60!}{29! \cdot 30!}$$

$$S = \frac{30}{2} \cdot \frac{60!}{30! \cdot 30!} = 15 \cdot \frac{60!}{(30!)^2}$$

$$\alpha = 15$$

Explanation

Consider the given series

$$\binom{n}{0}+\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}+\cdots+\frac{1}{n}\binom{n}{n-1}+\frac{1}{n+1}\binom{n}{n}=\frac{1023}{10}$$

Observe that each term has the form

$$\frac{1}{r+1}\binom{n}{r}$$

Use the identity

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

Applying this identity to every term,

$$LHS=\frac{1}{n+1}\left[\binom{n+1}{1}+\binom{n+1}{2}+\cdots+\binom{n+1}{n+1}\right]$$

Now use the binomial identity

$$\sum_{r=0}^{n+1}\binom{n+1}{r}=2^{,n+1}$$

Therefore,

$$\binom{n+1}{1}+\binom{n+1}{2}+\cdots+\binom{n+1}{n+1}=2^{,n+1}-1$$

Hence,

$$LHS=\frac{2^{,n+1}-1}{n+1}$$

Given that

$$\frac{2^{,n+1}-1}{n+1}=\frac{1023}{10}$$

Comparing denominators suggests

$$n+1=10$$

Substituting,

$$2^{10}-1=1024-1=1023$$

which matches the numerator exactly.

Therefore,

$$n+1=10$$

$$n=9$$

Hence, the correct answer is

$$\boxed{9}$$

To determine 2p + 3q, consider the expansion of (1+x)p (1−x)q up to the x² term, given that the coefficient of x is 4 and that of x² is −5.

Using the binomial theorem gives $$(1+x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \cdots$$ $$(1-x)^q = 1 - qx + \frac{q(q-1)}{2}x^2 + \cdots$$ Multiplying these series and collecting terms up to x² leads to an x-coefficient of $$p \cdot 1 + 1 \cdot (-q) = p - q,$$ so $$p - q = 4.$$

The coefficient of x² arises from three contributions: $$\frac{p(p-1)}{2} \cdot 1 + p \cdot (-q) + 1 \cdot \frac{q(q-1)}{2} = \frac{p(p-1)}{2} - pq + \frac{q(q-1)}{2} = \frac{p^2 - p + q^2 - q - 2pq}{2} = \frac{(p-q)^2 - (p+q)}{2},$$ and this must equal −5: $$\frac{(p-q)^2 - (p+q)}{2} = -5.$$

Since p − q = 4 implies (p − q)² = 16, substitution gives $$\frac{16 - (p+q)}{2} = -5 \quad\Longrightarrow\quad 16 - (p+q) = -10 \quad\Longrightarrow\quad p + q = 26.$$

Solving the system $$p - q = 4$$ $$p + q = 26$$ by addition gives 2p = 30, hence p = 15, and by subtraction gives 2q = 22, hence q = 11. Substituting these into 2p + 3q yields $$2p + 3q = 2(15) + 3(11) = 30 + 33 = 63.$$

The correct answer is Option 4: 63.

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

$$p_0 = a^{10}$$

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: $$12^3 - 12$$.

Given the binomial expansion of $$\left(\frac{4x}{5} - \frac{5}{2x}\right)^{2022}$$. The 1011th term from the end is 1024 times the 1011th term from the beginning.

Identify the terms.

In an expansion $$(a+b)^n$$ with $$n+1 = 2023$$ terms, the $$k$$th term from the end is the $$(n+2-k)$$th term from the beginning.

1011th term from the end = $$(2022 + 2 - 1011)$$th = 1013th term from the beginning.

Write the general term.

$$T_{r+1} = \binom{2022}{r}\left(\frac{4x}{5}\right)^{2022-r}\left(-\frac{5}{2x}\right)^r$$

1011th from beginning: $$T_{1011}$$ with $$r = 1010$$.

1013th from beginning: $$T_{1013}$$ with $$r = 1012$$.

Compute the ratio $$|T_{1013}|/|T_{1011}|$$.

$$\frac{|T_{1013}|}{|T_{1011}|} = \frac{\binom{2022}{1012}}{\binom{2022}{1010}} \cdot \left(\frac{4|x|}{5}\right)^{2022-1012-(2022-1010)} \cdot \left(\frac{5}{2|x|}\right)^{1012-1010}$$

Since $$1012 + 1010 = 2022$$, we have $$\binom{2022}{1012} = \binom{2022}{1010}$$, so their ratio is 1.

The remaining ratio simplifies to:

$$\left(\frac{4|x|}{5}\right)^{-2} \cdot \left(\frac{5}{2|x|}\right)^{2} = \frac{25}{16x^2} \cdot \frac{25}{4x^2} = \frac{625}{64x^4}$$

Solve for $$|x|$$.

$$\frac{625}{64x^4} = 1024$$

$$x^4 = \frac{625}{64 \times 1024} = \frac{625}{65536}$$

$$|x|^2 = \frac{25}{256}$$

$$|x| = \frac{5}{16}$$

Compute $$32|x|$$.

$$32|x| = 32 \times \frac{5}{16} = 10$$

The answer is Option B: $$10$$.

To find the fourth term coefficient, we first find $$n$$ using the property of consecutive binomial coefficients.

Let the three consecutive coefficients be $$^nC_{r-1}$$, $$^nC_r$$, and $$^nC_{r+1}$$.

• Ratio 1: $$\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r} = 5 \implies n - r + 1 = 5r \implies \mathbf{n - 6r = -1}$$
• Ratio 2: $$\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1} = \frac{20}{5} = 4 \implies n - r = 4r + 4 \implies \mathbf{n - 5r = 4}$$

Subtracting the equations:

$$(n - 5r) - (n - 6r) = 4 - (-1) \implies \mathbf{r = 5}$$

Substitute $$r=5$$ back: $$n - 5(5) = 4 \implies \mathbf{n = 29}$$

The fourth term in $$(1+x)^{29}$$ is $$^{29}C_3$$:

$$^{29}C_3 = \frac{29 \times 28 \times 27}{3 \times 2 \times 1} = 29 \times 14 \times 9 = \mathbf{3654}$$

The binomial expansion of $$(1-x)^{100}$$ is

$$(1-x)^{100}=\binom{100}{0}-\binom{100}{1}x+\binom{100}{2}x^2-\cdots+\binom{100}{100}x^{100}$$

The sum of the coefficients of the first 50 terms is

$$S=\binom{100}{0}-\binom{100}{1}+\binom{100}{2}-\cdots-\binom{100}{49}$$

The sum of all coefficients of the expansion can be obtained by putting $$x=1$$.

Therefore,

$$(1-1)^{100}=0$$

Hence,

$$\binom{100}{0}-\binom{100}{1}+\binom{100}{2}-\cdots+\binom{100}{100}=0$$

Since there are 101 terms, we split the expansion into:

• First 50 terms
• Middle term
• Last 50 terms

Thus,

$$S+\binom{100}{50}+S=0$$

Using the symmetry property $$\binom{n}{r}=\binom{n}{n-r}$$, the sum of the last 50 terms is equal to the sum of the first 50 terms.

Therefore,

$$2S+\binom{100}{50}=0$$

$$2S=-\binom{100}{50}$$

$$S=-\frac{1}{2}\binom{100}{50}$$

Now use the identity $$\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}$$ so that

$$\binom{100}{50}=\frac{100}{50}\binom{99}{49}$$

$$\binom{100}{50}=2\binom{99}{49}$$

Substituting,

$$S=-\frac{1}{2}\left(2\binom{99}{49}\right)$$

$$S=-\binom{99}{49}$$

Hence, the sum of the coefficients of the first 50 terms is

$$-\binom{99}{49}$$

Therefore, the correct answer is

$$-\binom{99}{49}$$

Three consecutive terms $$\binom{n+2}{r}$$, $$\binom{n+2}{r+1}$$, $$\binom{n+2}{r+2}$$ in the expansion of $$(1+x)^{n+2}$$ are in the ratio $$1:3:5$$.

From the first ratio:

$$\frac{\binom{n+2}{r+1}}{\binom{n+2}{r}} = \frac{n+2-r}{r+1} = 3$$

$$n + 2 - r = 3r + 3 \implies n = 4r + 1 \quad \cdots(1)$$

From the second ratio:

$$\frac{\binom{n+2}{r+2}}{\binom{n+2}{r+1}} = \frac{n+1-r}{r+2} = \frac{5}{3}$$

$$3(n+1-r) = 5(r+2) \implies 3n = 8r + 7 \quad \cdots(2)$$

Substituting (1) into (2):

$$3(4r+1) = 8r+7 \implies 12r+3 = 8r+7 \implies r = 1$$

$$n = 5$$, so the expansion is $$(1+x)^7$$.

The three consecutive terms:

$$\binom{7}{1} = 7, \quad \binom{7}{2} = 21, \quad \binom{7}{3} = 35$$

Ratio: $$7:21:35 = 1:3:5$$ ✓

Sum = $$7 + 21 + 35 = 63$$

The answer is Option B: $$63$$.

$$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$$

The expansion is $$(x - 3x^{-2})^n$$. The general term is:

$$T_{r+1} = \binom{n}{r} (x)^{n-r} (-3x^{-2})^r = \binom{n}{r} (-3)^r x^{n-r-2r} = \binom{n}{r} (-3)^r x^{n-3r}$$

$$T_1 (r=0) = \binom{n}{0} (-3)^0 = 1$$

$$T_2 (r=1) = \binom{n}{1} (-3)^1 = -3n$$

$$T_3 (r=2) = \binom{n}{2} (-3)^2 = 9 \frac{n(n-1)}{2}$$

$$1 - 3n + \frac{9n(n-1)}{2} = 376$$

$$9n^2 - 15n - 750 = 0$$

Since $$n \in N$$, we take the positive root: $$n = 10$$.

$$10 - 3r = 4 \implies 3r = 6 \implies r = 2$$ (we need the term where the power of $$x$$ is 4 $$n - 3r = 4$$)

The coefficient is $$\binom{n}{r} (-3)^r$$: $$\text{Coeff} = \binom{10}{2} (-3)^2$$

$$\text{Coeff} = \frac{10 \times 9}{2} \times 9 = 45 \times 9 = 405$$

We consider the expansion of $$\left(x^{2/3} + \frac{\alpha}{x^3}\right)^{22}$$ and note that its general term is $$T_{r+1} = \binom{22}{r} \left(x^{2/3}\right)^{22-r} \left(\frac{\alpha}{x^3}\right)^r = \binom{22}{r} \alpha^r \cdot x^{\frac{2(22-r)}{3} - 3r}$$.

The exponent of $$x$$ in this term is $$\frac{44 - 2r}{3} - 3r = \frac{44 - 11r}{3}$$. Setting this equal to zero yields $$44 - 11r = 0 \Rightarrow r = 4$$, so the term independent of $$x$$ is $$T_5 = \binom{22}{4} \alpha^4$$.

Computing $$\binom{22}{4}$$ gives $$\binom{22}{4} = \frac{22 \times 21 \times 20 \times 19}{4!} = \frac{175560}{24} = 7315$$, hence the constant term is $$7315 \alpha^4$$. Since this equals 7315, we have $$7315 \cdot \alpha^4 = 7315 \Rightarrow \alpha^4 = 1 \Rightarrow |\alpha| = 1$$.

We need to find the constant term in the expansion of $$\left(2x + \frac{1}{x^7} + 3x^2\right)^5$$.

Using the multinomial theorem, the general term is:

$$\frac{5!}{a!\, b!\, c!} (2x)^a \left(\frac{1}{x^7}\right)^b (3x^2)^c = \frac{5!}{a!\, b!\, c!} \cdot 2^a \cdot 3^c \cdot x^{a - 7b + 2c}$$

where $$a + b + c = 5$$ and $$a, b, c \geq 0$$.

For the constant term, the power of $$x$$ must be zero:

$$a - 7b + 2c = 0$$

We solve the system: $$a + b + c = 5$$ and $$a - 7b + 2c = 0$$.

From the first equation: $$a = 5 - b - c$$. Substituting into the second:

$$5 - b - c - 7b + 2c = 0 \implies 5 - 8b + c = 0 \implies c = 8b - 5$$

For non-negative integers: $$c \geq 0$$ requires $$b \geq 1$$, and $$a = 10 - 9b \geq 0$$ requires $$b \leq 1$$.

So $$b = 1$$, giving $$c = 3$$ and $$a = 1$$.

The constant term is:

$$\frac{5!}{1!\, 1!\, 3!} \cdot 2^1 \cdot 3^3 = \frac{120}{6} \cdot 2 \cdot 27 = 20 \times 54 = 1080$$

The answer is $$\boxed{1080}$$.

To find $$[\alpha]$$, we need to determine the constant term (the term where the power of $$x$$ is $$0$$) in the expansion of $$(3x^2 - \frac{1}{2x^5})^7$$.

The general term $$T_{r+1}$$ in the expansion of $$(a + b)^n$$ is $$^nC_r \cdot a^{n-r} \cdot b^r$$.

For $$(3x^2 - \frac{1}{2x^5})^7$$:

$$T_{r+1} = ^7C_r \cdot (3x^2)^{7-r} \cdot \left(-\frac{1}{2x^5}\right)^r$$

Combine the $$x$$ terms to find the total exponent:

$$T_{r+1} = ^7C_r \cdot 3^{7-r} \cdot \left(-\frac{1}{2}\right)^r \cdot \frac{x^{2(7-r)}}{x^{5r}}$$

$$x \text{ exponent} = 14 - 2r - 5r = 14 - 7r$$

Set the exponent to $$0$$:

$$14 - 7r = 0 \implies \mathbf{r = 2}$$

Plug $$r = 2$$ back into the coefficient part:

$$\alpha = ^7C_2 \cdot 3^{7-2} \cdot \left(-\frac{1}{2}\right)^2$$

$$\alpha = 21 \cdot 3^5 \cdot \frac{1}{4}$$

$$\alpha = 21 \cdot 243 \cdot 0.25 = \frac{5103}{4} = \mathbf{1275.75}$$

The problem asks for $$[\alpha]$$, which is the greatest integer $$\le \alpha$$:

$$[1275.75] = \mathbf{1275}$$

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 $$0^2 + 2^2 = 0 + 4 = 4$$.

The general term of the expansion $$\left(x^4 - \frac{1}{x^3}\right)^{15}$$ is:

$$ T_{r+1} = \binom{15}{r}(x^4)^{15-r}\left(-\frac{1}{x^3}\right)^r = \binom{15}{r}(-1)^r x^{4(15-r)-3r} = \binom{15}{r}(-1)^r x^{60-7r} $$

For the coefficient of $$x^{18}$$:

$$ 60 - 7r = 18 \implies 7r = 42 \implies r = 6 $$

The coefficient is:

$$ \binom{15}{6}(-1)^6 = \binom{15}{6} = \frac{15!}{6! \cdot 9!} = 5005 $$

The coefficient of $$x^{18}$$ is 5005.

We need to find the smallest $$\alpha > 0$$ such that the expansion of $$\left(x^{2/3} + \frac{2}{x^3}\right)^{30}$$ has a term $$\beta x^{-\alpha}$$, $$\beta \in \mathbb{N}$$.

$$T_{r+1} = \binom{30}{r} \left(x^{2/3}\right)^{30-r} \left(\frac{2}{x^3}\right)^r = \binom{30}{r} 2^r x^{\frac{2(30-r)}{3} - 3r}$$

Power of $$x$$: $$\frac{2(30-r)}{3} - 3r = \frac{60 - 2r - 9r}{3} = \frac{60 - 11r}{3}$$

$$\frac{60 - 11r}{3} = -\alpha \Rightarrow \alpha = \frac{11r - 60}{3}$$

For $$\alpha > 0$$: $$11r > 60 \Rightarrow r > \frac{60}{11} \approx 5.45$$, so $$r \geq 6$$.

For $$\alpha$$ to be a positive integer (or positive value with $$\beta \in \mathbb{N}$$), we need $$11r - 60$$ divisible by 3.

$$11r - 60 \equiv 2r \pmod{3}$$. So we need $$r \equiv 0 \pmod{3}$$.

Smallest valid $$r$$: $$r = 6$$ (since $$6 > 5.45$$ and $$6 \equiv 0 \pmod{3}$$).

$$\alpha = \frac{66 - 60}{3} = 2$$

Check $$\beta$$: $$\beta = \binom{30}{6} \cdot 2^6$$, which is a natural number. ✓

The answer is $$\boxed{2}$$.

Three consecutive terms of $$(1 + 2x)^n$$ have coefficients in the ratio $$2 : 5 : 8$$. Since the general term is $$T_{r+1} = \binom{n}{r}(2x)^r = \binom{n}{r} \cdot 2^r \cdot x^r$$, the coefficient of $$x^r$$ is $$\binom{n}{r} \cdot 2^r$$.

For three consecutive terms with indices $$r, r+1, r+2$$, the ratio of the coefficient of the $$(r+1)$$th term to the $$r$$th term gives $$\frac{\binom{n}{r+1} \cdot 2^{r+1}}{\binom{n}{r} \cdot 2^r} = \frac{5}{2} \implies \frac{n-r}{r+1} \cdot 2 = \frac{5}{2} \implies \frac{n-r}{r+1} = \frac{5}{4},$$ which leads to $$4(n-r) = 5(r+1) \implies 4n = 9r + 5 \quad \cdots (i).$$

Next, the ratio of the coefficient of the $$(r+2)$$th term to the $$(r+1)$$th term yields $$\frac{\binom{n}{r+2} \cdot 2^{r+2}}{\binom{n}{r+1} \cdot 2^{r+1}} = \frac{8}{5} \implies \frac{n-r-1}{r+2} \cdot 2 = \frac{8}{5} \implies \frac{n-r-1}{r+2} = \frac{4}{5},$$ which gives $$5(n-r-1) = 4(r+2) \implies 5n = 9r + 13 \quad \cdots (ii).$$

Subtracting $$(i)$$ from $$(ii)$$ gives $$n = 8$$, and substituting back into $$(i)$$ yields $$32 = 9r + 5 \implies r = 3$$.

Therefore the middle term, corresponding to index $$r+1 = 4$$, has coefficient $$\binom{8}{4} \cdot 2^4 = 70 \times 16 = 1120.$$ The correct answer is 1120.

The general term in the expansion of $$\left(3^{1/2} + 5^{1/4}\right)^{680}$$ is:

$$T_{r+1} = \binom{680}{r} \cdot 3^{(680-r)/2} \cdot 5^{r/4}$$

For the term to be an integer, both exponents must be non-negative integers:

Condition 1: $$\frac{680-r}{2}$$ is a non-negative integer $$\Rightarrow r$$ is even

Condition 2: $$\frac{r}{4}$$ is a non-negative integer $$\Rightarrow r$$ is divisible by 4

Both conditions are satisfied when $$r$$ is a multiple of 4.

$$r = 0, 4, 8, 12, \ldots, 680$$

Number of values = $$\frac{680}{4} + 1 = 170 + 1 = 171$$

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.

Since the coefficient of $$x^9$$ in $$\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$$ equals the coefficient of $$x^{-9}$$ in $$\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}$$, we start by finding the former. Considering the general term in the expansion of $$\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$$:
$$T_{r+1} = \binom{11}{r}(\alpha x^3)^{11-r}\left(\frac{1}{\beta x}\right)^r = \binom{11}{r}\frac{\alpha^{11-r}}{\beta^r} x^{33-4r}.$$
Substituting $$33 - 4r = 9$$ gives $$r = 6$$, so the coefficient of $$x^9$$ is $$\binom{11}{6}\frac{\alpha^5}{\beta^6}$$.

Next, in the expansion of $$\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}$$, the general term is:
$$T_{r+1} = \binom{11}{r}(\alpha x)^{11-r}\left(\frac{-1}{\beta x^3}\right)^r = \binom{11}{r}\frac{(-1)^r\alpha^{11-r}}{\beta^r} x^{11-4r}.$$
Requiring $$11 - 4r = -9$$ yields $$r = 5$$, and hence the coefficient of $$x^{-9}$$ is $$\binom{11}{5}\frac{(-1)^5\alpha^6}{\beta^5} = -\binom{11}{5}\frac{\alpha^6}{\beta^5}\,.$$

From the above, equating these coefficients and noting that $$\binom{11}{6} = \binom{11}{5} = 462$$ gives
$$\frac{\alpha^5}{\beta^6} = -\frac{\alpha^6}{\beta^5}\,,\quad\text{so}\quad \frac{1}{\beta} = -\alpha\,,\quad\text{and thus}\quad \alpha\beta = -1\,. $$

Therefore,
$$(\alpha\beta)^2 = (-1)^2 = 1\,, $$ so the correct answer is 1.

We need to find $$\alpha$$ where $$\sum_{r=0}^{2023} r^2 \binom{2023}{r} = 2023 \times \alpha \times 2^{2022}$$.

We know that $$\sum_{r=0}^{n} r\binom{n}{r} = n \cdot 2^{n-1}$$ and that $$\sum_{r=0}^{n} r^2\binom{n}{r} = n(n+1) \cdot 2^{n-2}$$. The latter follows from the identity $$r^2 = r(r-1) + r$$ and linearity of summation.

Indeed, by writing $$\sum r^2\binom{n}{r} = \sum r(r-1)\binom{n}{r} + \sum r\binom{n}{r}$$ we obtain $$n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1} = n(n-1) \cdot 2^{n-2} + 2n \cdot 2^{n-2} = n(n+1) \cdot 2^{n-2}$$.

Substituting $$n = 2023$$ gives $$\sum_{r=0}^{2023} r^2\binom{2023}{r} = 2023 \times 2024 \times 2^{2021}$$.

Comparing this with $$2023 \times \alpha \times 2^{2022}$$ leads to the equation $$2023 \times 2024 \times 2^{2021} = 2023 \times \alpha \times 2^{2022}\,$$ which simplifies to $$2024 \times 2^{2021} = \alpha \times 2^{2022}$$ and hence $$\alpha = \frac{2024}{2} = 1012$$.

The answer is 1012.

Find coefficient of $$x^7$$ in $$(1 - x + 2x^3)^{10}$$.

Using multinomial expansion: $$(1 - x + 2x^3)^{10} = \sum \frac{10!}{a!b!c!}(1)^a(-x)^b(2x^3)^c$$

where $$a + b + c = 10$$ and $$b + 3c = 7$$.

From $$b + 3c = 7$$: possible $$(c, b) = (0,7), (1,4), (2,1)$$

Case 1: $$c=0, b=7, a=3$$: $$\frac{10!}{3!7!0!}(-1)^7 = -120$$

Case 2: $$c=1, b=4, a=5$$: $$\frac{10!}{5!4!1!}(-1)^4 \cdot 2 = \frac{10!}{5!4!} \cdot 2 = 1260 \times 2 = 2520$$. Wait: $$\frac{10!}{5!4!1!} = \frac{3628800}{120 \cdot 24} = 1260$$. So $$1260 \times 2 = 2520$$.

Case 3: $$c=2, b=1, a=7$$: $$\frac{10!}{7!1!2!}(-1)^1 \cdot 4 = -\frac{10!}{7!2!} \times 4 = -360 \times 4$$. $$\frac{10!}{7!2!} = \frac{3628800}{5040 \times 2} = 360$$. So $$-360 \times 4 = -1440$$.

Total = $$-120 + 2520 - 1440 = 960$$

The answer is 960.

The binomial expansion of $$(2 + x)^9$$ is:

$$(2+x)^9 = \sum_{r=0}^{9}\binom{9}{r}2^{9-r}x^r$$

We need the mean of the coefficients of $$x, x^2, \ldots, x^7$$.

Sum of all coefficients (put $$x = 1$$): $$(2+1)^9 = 3^9 = 19683$$

Coefficient of $$x^0 = \binom{9}{0}2^9 = 512$$

Coefficient of $$x^8 = \binom{9}{8}2^1 = 18$$

Coefficient of $$x^9 = \binom{9}{9}2^0 = 1$$

Sum of coefficients from $$x^1$$ to $$x^7$$:

$$= 19683 - 512 - 18 - 1 = 19152$$

Mean = $$\frac{19152}{7} = 2736$$

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: $$2^{2n} - \binom{2n}{n}$$.

We need to evaluate $$\sum_{k=1}^{31} \binom{31}{k}\binom{31}{k-1} - \sum_{k=1}^{30} \binom{30}{k}\binom{30}{k-1}$$ and express it as $$\frac{\alpha \cdot 60!}{30! \cdot 31!}$$, then find $$16\alpha$$.

Noting that $$\binom{31}{k-1} = \binom{31}{32-k}$$ and applying the Vandermonde identity $$\sum_{k} \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$$ with $$m=n=31$$ and $$r=32$$, we obtain

$$\sum_{k=1}^{31} \binom{31}{k}\binom{31}{32-k} = \binom{62}{32}\,. $$

Similarly, since $$\binom{30}{k-1} = \binom{30}{31-k}$$ and using Vandermonde with $$m=n=30$$ and $$r=31$$, we get

$$\sum_{k=1}^{30} \binom{30}{k}\binom{30}{31-k} = \binom{60}{31}\,. $$

Therefore the left-hand side simplifies to

$$\binom{62}{32} - \binom{60}{31}\,. $$

We set this equal to $$\frac{\alpha \cdot 60!}{30! \cdot 31!}$$, so that

$$\alpha = \bigl(\binom{62}{32} - \binom{60}{31}\bigr)\frac{30! \cdot 31!}{60!}\,.$$

For the first term,

$$\binom{62}{32}\times\frac{30! \cdot 31!}{60!} =\frac{62!}{32! \cdot 30!}\times\frac{30! \cdot 31!}{60!} =\frac{62! \cdot 31!}{32! \cdot 60!} =\frac{62 \cdot 61}{32} =\frac{3782}{32} =\frac{1891}{16}\,. $$

For the second term,

$$\binom{60}{31}\times\frac{30! \cdot 31!}{60!} =\frac{60!}{31! \cdot 29!}\times\frac{30! \cdot 31!}{60!} =\frac{30!}{29!} =30\,. $$

Hence

$$\alpha = \frac{1891}{16} - 30 = \frac{1891 - 480}{16} = \frac{1411}{16},$$

so

$$16\alpha = 1411\,. $$

Therefore, the correct answer is Option A: 1411.

We are given: $$9^n - 8n - 1 = 64\alpha$$ and $$6^n - 5n - 1 = 25\beta$$, for $$n \geq 5$$. Expanding $$9^n = (1+8)^n$$ by the binomial theorem yields $$9^n = \sum_{k=0}^{n} \binom{n}{k} 8^k = 1 + 8n + \binom{n}{2}8^2 + \binom{n}{3}8^3 + \ldots + 8^n,$$ so $$9^n - 8n - 1 = \binom{n}{2}8^2 + \binom{n}{3}8^3 + \ldots + \binom{n}{n}8^n = 64\Bigl[\binom{n}{2} + \binom{n}{3}\cdot 8 + \binom{n}{4}\cdot 8^2 + \ldots + \binom{n}{n}\cdot 8^{n-2}\Bigr],$$ and therefore $$\alpha = \sum_{k=2}^{n}\binom{n}{k}\cdot 8^{k-2}.$$ Similarly, expanding $$6^n = (1+5)^n$$ gives $$6^n - 5n - 1 = \binom{n}{2}5^2 + \binom{n}{3}5^3 + \ldots + \binom{n}{n}5^n = 25\Bigl[\binom{n}{2} + \binom{n}{3}\cdot 5 + \binom{n}{4}\cdot 5^2 + \ldots + \binom{n}{n}\cdot 5^{n-2}\Bigr],$$ so $$\beta = \sum_{k=2}^{n}\binom{n}{k}\cdot 5^{k-2}.$$ Computing $$\alpha - \beta$$ we have $$\alpha - \beta = \sum_{k=2}^{n}\binom{n}{k}(8^{k-2} - 5^{k-2}).$$ Since for $$k = 2$$ the term is zero, the sum effectively starts from $$k = 3$$: $$\alpha - \beta = \sum_{k=3}^{n}\binom{n}{k}(8^{k-2} - 5^{k-2}) = \binom{n}{3}(8^1 - 5^1) + \binom{n}{4}(8^2 - 5^2) + \ldots + \binom{n}{n}(8^{n-2} - 5^{n-2}).$$ This matches Option C: $$\binom{n}{3}(8-5) + \binom{n}{4}(8^2-5^2) + \ldots + \binom{n}{n}(8^{n-2}-5^{n-2}).$$ Therefore, the answer is Option C.

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 need the term independent of $$x$$ in:

$$\left(1 - x^2 + 3x^3\right)\left(\frac{5}{2}x^3 - \frac{1}{5x^2}\right)^{11}$$

Find the general term of the binomial expansion:

The general term of $$\left(\frac{5}{2}x^3 - \frac{1}{5x^2}\right)^{11}$$ is:

$$T_{r+1} = \binom{11}{r}\left(\frac{5}{2}\right)^{11-r}\left(-\frac{1}{5}\right)^r x^{3(11-r) - 2r} = \binom{11}{r}\left(\frac{5}{2}\right)^{11-r}\left(-\frac{1}{5}\right)^r x^{33-5r}$$

Identify which terms contribute to the constant term in the product:

When multiplied by $$(1 - x^2 + 3x^3)$$, we need:

• From $$1 \cdot T_{r+1}$$: need $$33 - 5r = 0 \Rightarrow r = 33/5$$ (not an integer, no contribution)
• From $$(-x^2) \cdot T_{r+1}$$: need $$33 - 5r + 2 = 0 \Rightarrow 5r = 35 \Rightarrow r = 7$$ $$\checkmark$$
• From $$(3x^3) \cdot T_{r+1}$$: need $$33 - 5r + 3 = 0 \Rightarrow 5r = 36 \Rightarrow r = 36/5$$ (not an integer, no contribution)

Compute the coefficient for $$r = 7$$:

$$T_8 = \binom{11}{7}\left(\frac{5}{2}\right)^{4}\left(-\frac{1}{5}\right)^{7}$$

$$= 330 \times \frac{625}{16} \times \frac{(-1)}{78125}$$

$$= 330 \times \frac{-625}{16 \times 78125}$$

$$= 330 \times \frac{-1}{16 \times 125}$$

$$= \frac{-330}{2000} = \frac{-33}{200}$$

Multiply by the coefficient $$(-1)$$ from the $$(-x^2)$$ term:

$$\text{Constant term} = (-1) \times \frac{-33}{200} = \frac{33}{200}$$

The correct answer is Option B: $$\dfrac{33}{200}$$.

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 find the sum of all possible integral values of $$n$$ such that the sum of coefficients of all positive powers of $$x$$ in $$\left(x^n + \frac{2}{x^5}\right)^7$$ is 939.

The general term of the expansion is $$T_{r+1} = \binom{7}{r}(x^n)^{7-r}\left(\frac{2}{x^5}\right)^r = \binom{7}{r} \cdot 2^r \cdot x^{n(7-r) - 5r}$$, so the power of $$x$$ in this term is $$7n - r(n + 5)$$. For this exponent to be positive we require $$7n - r(n + 5) > 0 \implies r < \frac{7n}{n + 5}$$.

Since the coefficient of $$T_{r+1}$$ is $$\binom{7}{r} \cdot 2^r$$, we need $$\sum_{\substack{r=0 \\ 7n - r(n+5) > 0}}^{7} \binom{7}{r} \cdot 2^r = 939$$. Computing gives $$\sum_{r=0}^{4} \binom{7}{r} \cdot 2^r = 1 + 14 + 84 + 280 + 560 = 939$$, so exactly $$r = 0, 1, 2, 3, 4$$ produce positive powers and $$r = 5$$ is the first to give a non-positive power.

Hence for $$r = 4$$ we need $$7n - 4(n + 5) = 3n - 20 > 0 \implies n > \frac{20}{3} \implies n \geq 7$$, while for $$r = 5$$ we require $$7n - 5(n + 5) = 2n - 25 \leq 0 \implies n \leq 12$$ (when $$n = 12$$: $$2(12) - 25 = -1 < 0$$ $$\checkmark$$; when $$n = 13$$: $$2(13) - 25 = 1 > 0$$ $$\times$$). Therefore $$n \in \{7, 8, 9, 10, 11, 12\}$$.

Summing these values yields $$7 + 8 + 9 + 10 + 11 + 12 = 57$$.

The correct answer is $$\boxed{57}$$.

We need to find the least value $$n_0$$ such that the 9th term in the binomial expansion of $$(3 + 6x)^n$$ (in increasing powers of $$6x$$) is greatest when $$x = \frac{3}{2}$$, and then compute $$k + n_0$$ where $$k$$ is the ratio of the coefficient of $$x^6$$ to the coefficient of $$x^3$$.

Since the $$(r+1)$$-th term in the expansion of $$(3 + 6x)^n$$ is given by $$T_{r+1} = \binom{n}{r} 3^{n-r} (6x)^r,$$ substituting $$x = \frac{3}{2}$$ yields $$T_{r+1} = \binom{n}{r} 3^{n-r} \cdot 6^r \cdot \left(\frac{3}{2}\right)^r = \binom{n}{r} 3^{n-r} \cdot \frac{6^r \cdot 3^r}{2^r} = \binom{n}{r} 3^n \cdot \frac{6^r}{2^r} = \binom{n}{r} 3^n \cdot 3^r = \binom{n}{r} 3^{n+r}.$$

From this, the ratio of consecutive terms is $$\frac{T_{r+1}}{T_r} = \frac{\binom{n}{r} 3^{n+r}}{\binom{n}{r-1} 3^{n+r-1}} = \frac{n-r+1}{r} \cdot 3.$$ For the 9th term (corresponding to $$r = 8$$) to be the greatest, we require both $$T_9 \ge T_8$$ and $$T_9 \ge T_{10}$$.

From $$\frac{T_9}{T_8} \ge 1 \quad(\text{ratio at } r = 8),$$ we have $$\frac{3(n - 7)}{8} \ge 1 \implies n \ge \frac{29}{3} \approx 9.67 \implies n \ge 10.$$

Similarly, from $$\frac{T_{10}}{T_9} \le 1 \quad(\text{ratio at } r = 9),$$ we get $$\frac{3(n - 8)}{9} \le 1 \implies n - 8 \le 3 \implies n \le 11.$$ Hence $$10 \le n \le 11$$ and the least value satisfying these inequalities is $$n_0 = 10$$.

Next, to determine $$k$$, the ratio of the coefficient of $$x^6$$ to that of $$x^3$$ in $$(3 + 6x)^{10},$$ note that the coefficient of $$x^r$$ is $$\binom{10}{r} 3^{10-r} \cdot 6^r.$$ Therefore, $$\text{Coefficient of }x^6 = \binom{10}{6} \cdot 3^4 \cdot 6^6,$$ $$\text{Coefficient of }x^3 = \binom{10}{3} \cdot 3^7 \cdot 6^3,$$ and hence $$k = \frac{\binom{10}{6} \cdot 3^4 \cdot 6^6}{\binom{10}{3} \cdot 3^7 \cdot 6^3} = \frac{\binom{10}{6}}{\binom{10}{3}} \cdot \frac{6^3}{3^3} = \frac{210}{120} \cdot \frac{216}{27} = \frac{7}{4} \cdot 8 = 14.$$

Finally, combining these results gives $$k + n_0 = 14 + 10 = 24,$$ so the answer is $$\boxed{24}$$.

We need to find $$L$$ such that $$\displaystyle\sum_{k=1}^{10} k^2 \binom{10}{k}^2 = 22000L$$.

We begin by simplifying $$k^2\binom{10}{k}$$ using the standard identity $$k\binom{n}{k} = n\binom{n-1}{k-1}$$. Applying this once gives $$k\binom{10}{k} = 10\binom{9}{k-1}$$, and then multiplying by $$k$$ and writing $$k = (k-1) + 1$$:

$$k^2\binom{10}{k} = k \cdot 10\binom{9}{k-1} = 10\bigl[(k-1) + 1\bigr]\binom{9}{k-1} = 10(k-1)\binom{9}{k-1} + 10\binom{9}{k-1}.$$

Now applying the identity again to the first term: $$(k-1)\binom{9}{k-1} = 9\binom{8}{k-2}$$. So we obtain $$k^2\binom{10}{k} = 90\binom{8}{k-2} + 10\binom{9}{k-1}$$.

Substituting into the sum and using a second copy of $$\binom{10}{k}$$:

$$\sum_{k=1}^{10} k^2\binom{10}{k}^2 = 90\sum_{k=2}^{10}\binom{8}{k-2}\binom{10}{k} + 10\sum_{k=1}^{10}\binom{9}{k-1}\binom{10}{k}.$$

For the first sum, we substitute $$j = k - 2$$: $$\displaystyle\sum_{j=0}^{8}\binom{8}{j}\binom{10}{j+2}$$. Using the identity $$\binom{10}{j+2} = \binom{10}{8-j}$$ and the Vandermonde convolution $$\displaystyle\sum_{j=0}^{8}\binom{8}{j}\binom{10}{8-j} = \binom{18}{8}$$. We compute $$\binom{18}{8} = 43758$$.

For the second sum, we substitute $$j = k - 1$$: $$\displaystyle\sum_{j=0}^{9}\binom{9}{j}\binom{10}{j+1}$$. Since $$\binom{10}{j+1} = \binom{10}{9-j}$$, the Vandermonde convolution gives $$\displaystyle\sum_{j=0}^{9}\binom{9}{j}\binom{10}{9-j} = \binom{19}{9} = 92378$$.

Combining: $$\displaystyle\sum_{k=1}^{10} k^2\binom{10}{k}^2 = 90 \times 43758 + 10 \times 92378 = 3{,}938{,}220 + 923{,}780 = 4{,}862{,}000$$.

Setting $$4{,}862{,}000 = 22000L$$, we get $$L = \frac{4{,}862{,}000}{22{,}000} = 221$$.

Hence, the correct answer is $$\boxed{221}$$.

Find the coefficient of $$x^{10}$$ in $$\left(\frac{\sqrt{x}}{5^{1/4}} + \frac{\sqrt{5}}{x^{1/3}}\right)^{60}$$, expressed as $$5^k \cdot l$$ where $$\gcd(l, 5) = 1$$.

First, we write the general term in the binomial expansion. The $$(r+1)$$th term is given by $$T_{r+1} = \binom{60}{r} \left(\frac{\sqrt{x}}{5^{1/4}}\right)^{60-r} \left(\frac{\sqrt{5}}{x^{1/3}}\right)^r$$. Substituting the powers yields $$= \binom{60}{r} \cdot \frac{x^{(60-r)/2}}{5^{(60-r)/4}} \cdot \frac{5^{r/2}}{x^{r/3}}$$ which simplifies to $$= \binom{60}{r} \cdot 5^{r/2 - (60-r)/4} \cdot x^{(60-r)/2 - r/3}$$.

Next, we determine the value of $$r$$ that makes the power of $$x$$ equal to 10. Setting $$\frac{60-r}{2} - \frac{r}{3} = 10$$ leads to $$\frac{3(60-r) - 2r}{6} = 10$$, then $$180 - 3r - 2r = 60$$, so $$5r = 120$$ and hence $$r = 24$$.

Now, the exponent of 5 in the coefficient is $$\frac{r}{2} - \frac{60-r}{4}$$. Substituting $$r=24$$ gives $$\frac{24}{2} - \frac{36}{4} = 12 - 9 = 3$$.

Therefore, the coefficient has the form $$\binom{60}{24} \cdot 5^3$$, so we need to find the power of 5 dividing $$\binom{60}{24}$$. Using Legendre's formula, the exponent of 5 in $$n!$$ is $$\sum_{i=1}^{\infty} \lfloor n/5^i \rfloor$$. Thus, the power of 5 in $$60!$$ is $$\lfloor 60/5 \rfloor + \lfloor 60/25 \rfloor + \lfloor 60/125 \rfloor = 12 + 2 + 0 = 14$$, in $$24!$$ it is $$\lfloor 24/5 \rfloor + \lfloor 24/25 \rfloor = 4 + 0 = 4$$, and in $$36!$$ it is $$\lfloor 36/5 \rfloor + \lfloor 36/25 \rfloor = 7 + 1 = 8$$. Hence, the power of 5 in $$\binom{60}{24}$$ equals $$14 - 4 - 8 = 2$$.

Adding the exponent from the binomial coefficient to the exponent obtained earlier gives the total power of 5 as $$k = 3 + 2 = 5$$.

The answer is $$\boxed{5}$$.

We need to find coefficients of $$x^{-1}$$ and $$x^{-3}$$ in $$\left(2x^{1/5} - \frac{1}{x^{1/5}}\right)^{15}$$.

$$T_{r+1} = \binom{15}{r}(2x^{1/5})^{15-r}\left(-\frac{1}{x^{1/5}}\right)^r = \binom{15}{r}2^{15-r}(-1)^r x^{(15-r)/5} \cdot x^{-r/5} = \binom{15}{r}2^{15-r}(-1)^r x^{(15-2r)/5}$$

Setting $$\frac{15-2r}{5} = -1$$ gives $$15 - 2r = -5 \Rightarrow r = 10$$, and hence $$m = \binom{15}{10}2^5(-1)^{10} = \binom{15}{5} \cdot 32 = 3003 \times 32 = 96096$$.

Similarly, $$\frac{15-2r}{5} = -3$$ leads to $$15 - 2r = -15 \Rightarrow r = 15$$, and $$n = \binom{15}{15}2^0(-1)^{15} = -1$$.

Therefore, $$mn^2 = 96096 \times 1 = 96096$$, and $$96096 = \binom{15}{5} \times 32 = 3003 \times 32$$.

We then solve $$\binom{15}{r} \cdot 2^r = 3003 \times 32 = \binom{15}{5} \times 2^5$$, giving $$r = 5$$.

Therefore, the value of $$r = \textbf{5}$$.

In the expansion of $$\left(2x^3 + \frac{3}{x^k}\right)^{12}$$, the general term is:

$$T_{r+1} = \binom{12}{r} (2x^3)^{12-r} \left(\frac{3}{x^k}\right)^r = \binom{12}{r} 2^{12-r} \cdot 3^r \cdot x^{3(12-r) - kr}$$

For the constant term, the power of $$x$$ must be zero:

$$36 - 3r - kr = 0 \implies r(3 + k) = 36 \implies r = \frac{36}{3 + k}$$

Since $$0 \leq r \leq 12$$ and $$r$$ must be a positive integer, $$(3 + k)$$ must divide 36 and $$\frac{36}{3+k} \leq 12$$, meaning $$3 + k \geq 3$$, i.e., $$k \geq 0$$. Since $$k$$ is a positive integer, $$k \geq 1$$.

Divisors of 36 that are at least 4: $$4, 6, 9, 12, 18, 36$$.

So $$k \in \{1, 3, 6, 9, 15, 33\}$$ with $$r \in \{9, 6, 4, 3, 2, 1\}$$ respectively.

The constant term is $$\binom{12}{r} \cdot 2^{12-r} \cdot 3^r$$.

We need this to equal $$2^8 \cdot l$$ where $$l$$ is odd. This means the highest power of 2 dividing the constant term must be exactly 8.

The power of 2 in the constant term is $$(12 - r) + v_2\binom{12}{r})$$, where $$v_2$$ denotes the 2-adic valuation.

For each case:

$$k = 1, r = 9$$: $$v_2(2^3) + v_2\binom{12}{9} = 3 + v_2(220) = 3 + 2 = 5$$. Not 8.

$$k = 3, r = 6$$: $$v_2(2^6) + v_2\binom{12}{6} = 6 + v_2(924) = 6 + 2 = 8$$. Yes!

$$k = 6, r = 4$$: $$v_2(2^8) + v_2\binom{12}{4} = 8 + v_2(495) = 8 + 0 = 8$$. Yes!

$$k = 9, r = 3$$: $$v_2(2^9) + v_2\binom{12}{3} = 9 + v_2(220) = 9 + 2 = 11$$. Not 8.

$$k = 15, r = 2$$: $$v_2(2^{10}) + v_2\binom{12}{2} = 10 + v_2(66) = 10 + 1 = 11$$. Not 8.

$$k = 33, r = 1$$: $$v_2(2^{11}) + v_2\binom{12}{1} = 11 + v_2(12) = 11 + 2 = 13$$. Not 8.

The values of $$k$$ that work are $$k = 3$$ and $$k = 6$$.

Hence the answer is $$\boxed{2}$$.

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 $$139$$.

We need to find the coefficient of $$x^3$$ in $$(1+x)^p(1-x)^q$$, given that the coefficients of $$x$$ and $$x^2$$ are $$-3$$ and $$-5$$ respectively.

Since the coefficient of $$x$$ is $$-3$$, $$[x^1]: \binom{p}{1}(1) + \binom{q}{1}(-1) = p - q = -3 \quad \cdots (1)$$

Next, since the coefficient of $$x^2$$ is $$-5$$, $$[x^2]: \binom{p}{2} - pq + \binom{q}{2} = \dfrac{p(p-1)}{2} - pq + \dfrac{q(q-1)}{2} = -5$$

$$p^2 - p + q^2 - q - 2pq = -10$$

$$(p - q)^2 - (p + q) = -10$$

$$9 - (p + q) = -10$$

$$p + q = 19 \quad \cdots (2)$$

From (1) and (2), $$p = 8$$, $$q = 11$$. Check: $$p, q \le 15$$. $$\checkmark$$

Next, $$[x^3] = \sum_{j=0}^{3} \binom{p}{3-j}\binom{q}{j}(-1)^j$$

$$= \binom{8}{3} - \binom{8}{2}\binom{11}{1} + \binom{8}{1}\binom{11}{2} - \binom{11}{3}$$

$$= 56 - 28 \times 11 + 8 \times 55 - 165$$

$$= 56 - 308 + 440 - 165$$

$$= 23$$

The correct answer is $$\boxed{23}$$.

We need to find the maximum value of the term independent of $$t$$ in the expansion of $$\left(t^2 x^{1/5} + \dfrac{(1-x)^{1/10}}{t}\right)^{15}$$, $$x \ge 0$$.

Expanding by the binomial theorem, the general term can be written as:

$$T_{r+1} = \binom{15}{r} \left(t^2 x^{1/5}\right)^{15-r} \left(\dfrac{(1-x)^{1/10}}{t}\right)^r$$

$$= \binom{15}{r} t^{2(15-r)} x^{(15-r)/5} \cdot t^{-r} (1-x)^{r/10}$$

$$= \binom{15}{r} t^{30-2r-r} x^{(15-r)/5} (1-x)^{r/10}$$

$$= \binom{15}{r} t^{30-3r} x^{(15-r)/5} (1-x)^{r/10}$$

For this term to be independent of $$t$$, the exponent of $$t$$ must be zero, which implies

$$30 - 3r = 0 \implies r = 10$$

Substituting $$r = 10$$ into the general term yields the term free of $$t$$ as:

$$T_{11} = \binom{15}{10} x^{(15-10)/5} (1-x)^{10/10} = \binom{15}{5} x(1-x) = 3003 \cdot x(1-x)$$

Thus the problem reduces to maximizing the function $$x(1-x)$$ for $$x \ge 0$$. Applying the AM-GM inequality gives

$$x(1-x) \le \left(\dfrac{x + (1-x)}{2}\right)^2 = \dfrac{1}{4}$$, with equality when $$x = \dfrac{1}{2}$$.

Substituting this maximum into the expression for $$T_{11}$$ gives

$$K = 3003 \cdot \dfrac{1}{4} = \dfrac{3003}{4}$$

$$8K = 8 \cdot \dfrac{3003}{4} = 2 \cdot 3003 = 6006$$

Therefore, the maximum value of the term independent of $$t$$ in the expansion is $$6006$$.

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 need to find the middle term coefficients of three expansions and use them to form an A.P.

For $$\left(\dfrac{1}{\sqrt{6}} + \beta x\right)^4$$ (even power 4), the middle term is the 3rd term ($$r = 2$$):

$$\binom{4}{2}\left(\dfrac{1}{\sqrt{6}}\right)^2 (\beta x)^2 = 6 \cdot \dfrac{1}{6} \cdot \beta^2 x^2 = \beta^2 x^2$$

So the coefficient $$a_1 = \beta^2$$.

For $$(1 - 3\beta x)^2$$ (even power 2), the middle term is the 2nd term ($$r = 1$$):

$$\binom{2}{1}(1)(-3\beta x) = -6\beta x$$

So the coefficient $$a_2 = -6\beta$$.

For $$\left(1 - \dfrac{\beta}{2}x\right)^6$$ (even power 6), the middle term is the 4th term ($$r = 3$$):

$$\binom{6}{3}\left(-\dfrac{\beta}{2}\right)^3 x^3 = 20 \cdot \left(-\dfrac{\beta^3}{8}\right) x^3 = -\dfrac{5\beta^3}{2} x^3$$

So the coefficient $$a_3 = -\dfrac{5\beta^3}{2}$$.

Since $$a_1, a_2, a_3$$ form an A.P., we have $$2a_2 = a_1 + a_3$$:

$$2(-6\beta) = \beta^2 + \left(-\dfrac{5\beta^3}{2}\right)$$

$$-12\beta = \beta^2 - \dfrac{5\beta^3}{2}$$

Since $$\beta > 0$$, we can divide both sides by $$\beta$$:

$$-12 = \beta - \dfrac{5\beta^2}{2}$$

$$\dfrac{5\beta^2}{2} - \beta - 12 = 0$$

$$5\beta^2 - 2\beta - 24 = 0$$

Using the quadratic formula: $$\beta = \dfrac{2 \pm \sqrt{4 + 480}}{10} = \dfrac{2 \pm 22}{10}$$

Since $$\beta > 0$$, we get $$\beta = \dfrac{24}{10} = \dfrac{12}{5}$$.

Now we find the common difference $$d = a_2 - a_1 = -6\beta - \beta^2 = -6 \cdot \dfrac{12}{5} - \dfrac{144}{25} = -\dfrac{72}{5} - \dfrac{144}{25} = -\dfrac{360 + 144}{25} = -\dfrac{504}{25}$$.

Finally, $$50 - \dfrac{2d}{\beta^2} = 50 - \dfrac{2 \times \left(-\dfrac{504}{25}\right)}{\dfrac{144}{25}} = 50 - \dfrac{-\dfrac{1008}{25}}{\dfrac{144}{25}} = 50 + \dfrac{1008}{144} = 50 + 7 = 57$$.

Hence, the answer is $$57$$.

We have the binomial expansion of $$\left(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}}\right)^n$$ in increasing powers of $$\frac{1}{\sqrt[4]{3}}$$. The general term is $$T_{r+1} = \binom{n}{r}(\sqrt[4]{2})^{n-r}\left(\frac{1}{\sqrt[4]{3}}\right)^r = \binom{n}{r}\cdot 2^{(n-r)/4}\cdot 3^{-r/4}$$.

The fifth term from the beginning is $$T_5$$ (with $$r = 4$$): $$T_5 = \binom{n}{4}\cdot 2^{(n-4)/4}\cdot 3^{-1}$$.

The fifth term from the end is the $$(n+2-5)$$th = $$(n-3)$$th term from the beginning, which is $$T_{n-3}$$ (with $$r = n-4$$): $$T_{n-3} = \binom{n}{n-4}\cdot 2^{4/4}\cdot 3^{-(n-4)/4} = \binom{n}{4}\cdot 2\cdot 3^{-(n-4)/4}$$.

The ratio $$\frac{T_5}{T_{n-3}} = \frac{2^{(n-4)/4}\cdot 3^{-1}}{2\cdot 3^{-(n-4)/4}} = \frac{2^{(n-4)/4}}{2}\cdot\frac{3^{(n-4)/4}}{3} = \frac{2^{(n-4)/4}\cdot 3^{(n-4)/4}}{6} = \frac{6^{(n-4)/4}}{6}$$.

We are told this ratio equals $$\sqrt[4]{6} : 1 = 6^{1/4}$$. So $$\frac{6^{(n-4)/4}}{6} = 6^{1/4}$$, giving $$6^{(n-4)/4} = 6^{1+1/4} = 6^{5/4}$$.

Therefore $$\frac{n-4}{4} = \frac{5}{4}$$, so $$n - 4 = 5$$, hence $$n = 9$$.

Now we find the sixth term from the beginning, $$T_6$$ (with $$r = 5$$):

$$T_6 = \binom{9}{5}(\sqrt[4]{2})^{4}\left(\frac{1}{\sqrt[4]{3}}\right)^5 = \binom{9}{5}\cdot 2\cdot \frac{1}{3^{5/4}}$$

We compute $$\binom{9}{5} = \binom{9}{4} = 126$$. Also $$3^{5/4} = 3\cdot 3^{1/4} = 3\sqrt[4]{3}$$.

So $$T_6 = 126\cdot 2\cdot\frac{1}{3\sqrt[4]{3}} = \frac{252}{3\sqrt[4]{3}} = \frac{84}{\sqrt[4]{3}}$$.

We are told $$T_6 = \frac{\alpha}{\sqrt[4]{3}}$$, so $$\alpha = 84$$.

Hence, the correct answer is 84.

We need to evaluate $$1 + (2 + {}^{49}C_1 + {}^{49}C_2 + \ldots + {}^{49}C_{49})({}^{50}C_2 + {}^{50}C_4 + \ldots + {}^{50}C_{50})$$.

We know that $${}^{49}C_0 + {}^{49}C_1 + \ldots + {}^{49}C_{49} = 2^{49}$$, so $${}^{49}C_1 + {}^{49}C_2 + \ldots + {}^{49}C_{49} = 2^{49} - 1$$.

Therefore, $$2 + {}^{49}C_1 + \ldots + {}^{49}C_{49} = 2 + 2^{49} - 1 = 2^{49} + 1$$.

Now for the second factor: the sum of even-indexed binomial coefficients of $${}^{50}C_k$$. We know $${}^{50}C_0 + {}^{50}C_2 + {}^{50}C_4 + \ldots + {}^{50}C_{50} = 2^{49}$$. So $${}^{50}C_2 + {}^{50}C_4 + \ldots + {}^{50}C_{50} = 2^{49} - {}^{50}C_0 = 2^{49} - 1$$.

So the expression becomes:

$$1 + (2^{49} + 1)(2^{49} - 1) = 1 + 2^{98} - 1 = 2^{98}$$

So $$2^{98} = 2^n \cdot m$$ where $$m$$ is odd. This means $$n = 98$$ and $$m = 1$$.

Therefore $$n + m = 98 + 1 = 99$$.

Hence, the correct answer is 99.

First, we need to find $$m + n$$ where $$\binom{40}{0} + \binom{41}{1} + \binom{42}{2} + \cdots + \binom{60}{20} = \frac{m}{n} \cdot \binom{60}{20}$$ and $$\gcd(m, n) = 1$$.

Next, we apply the Vandermonde-type identity: $$\binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}$$. Rewriting each term gives $$\binom{40+k}{k} = \binom{40+k}{40}$$ for $$k = 0, 1, 2, \ldots, 20$$.

Substituting this result into the sum yields $$\sum_{k=0}^{20} \binom{40+k}{40} = \binom{61}{41} = \binom{61}{20}$$.

Now, we express $$\binom{61}{20}$$ as a multiple of $$\binom{60}{20}$$. Since $$\binom{61}{20} = \frac{61!}{20! \cdot 41!} = \frac{61}{41} \cdot \frac{60!}{20! \cdot 40!} = \frac{61}{41} \cdot \binom{60}{20}$$, we have $$\frac{m}{n} = \frac{61}{41}$$ and because 61 and 41 are both prime, $$\gcd(61, 41) = 1$$.

Therefore, $$m + n = 61 + 41 = 102$$, and hence the answer is $$102$$.

1. Write the general term of the expansion

The general term $$T_{r+1}$$ in the binomial expansion of $$(a + b)^n$$ is given by the formula:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

Substituting $$a = 2^{\frac{1}{3}}$$, $$b = 3^{\frac{1}{4}}$$, and $$n = 12$$, we get:

$$T_{r+1} = \binom{12}{r} \left(2^{\frac{1}{3}}\right)^{12-r} \left(3^{\frac{1}{4}}\right)^r$$

$$T_{r+1} = \binom{12}{r} 2^{\frac{12-r}{3}} 3^{\frac{r}{4}}$$

Here, the variable $$r$$ can take any integer value from $$0$$ to $$12$$.

2. Determine the condition for rational terms

For a term to be a rational number, the exponents of both base prime numbers $$2$$ and $$3$$ must be integers. This gives us two conditions:

The fraction $$\frac{12-r}{3}$$ must be an integer, which implies that $$12-r$$ is a multiple of $$3$$. Consequently, $$r$$ must be a multiple of $$3$$.

The fraction $$\frac{r}{4}$$ must be an integer, which implies that $$r$$ must be a multiple of $$4$$.

To satisfy both conditions simultaneously, $$r$$ must be a common multiple of both $$3$$ and $$4$$. The least common multiple is:

$$\text{LCM}(3, 4) = 12$$

3. Find the valid values of r

Since the range of $$r$$ is restricted to $$0 \leq r \leq 12$$, the only possible integer values for $$r$$ that are multiples of $$12$$ are:

$$r = 0$$

$$r = 12$$

4. Calculate the rational terms

Now, evaluate the value of the general term at these two specific values of $$r$$:

When $$r = 0$$:

$$T_1 = \binom{12}{0} 2^{\frac{12-0}{3}} 3^{\frac{0}{4}}$$

$$T_1 = 1 \times 2^4 \times 3^0$$

$$T_1 = 1 \times 16 \times 1 = 16$$

When $$r = 12$$:

$$T_{13} = \binom{12}{12} 2^{\frac{12-12}{3}} 3^{\frac{12}{4}}$$

$$T_{13} = 1 \times 2^0 \times 3^3$$

$$T_{13} = 1 \times 1 \times 27 = 27$$

5. Compute the sum of the rational terms

Add the two rational terms together to find the total sum:

$$\text{Sum} = T_1 + T_{13}$$

We are given $$(1-y)^m(1+y)^n = 1 + a_1 y + a_2 y^2 + \ldots$$ with $$a_1 = a_2 = 10$$.

Expanding to first and second order: $$(1-y)^m = 1 - my + \frac{m(m-1)}{2}y^2 - \ldots$$ $$(1+y)^n = 1 + ny + \frac{n(n-1)}{2}y^2 + \ldots$$

The coefficient of $$y$$ in the product is $$n - m = a_1 = 10$$, so $$n - m = 10$$.

The coefficient of $$y^2$$ is: $$\frac{n(n-1)}{2} - mn + \frac{m(m-1)}{2} = a_2 = 10.$$

Simplifying: $$\frac{n^2 - n - 2mn + m^2 - m}{2} = 10$$, which gives $$\frac{(n-m)^2 - (n+m)}{2} = 10$$ $$(n-m)^2 - (n+m) = 20$$ $$100 - (n+m) = 20$$ $$n + m = 80.$$

Therefore, $$m + n = \boxed{80}$$.

The general term in the expansion of $$\left(3^{1/4} + 5^{1/8}\right)^{60}$$ is given by $$T_{r+1} = \binom{60}{r} \cdot 3^{(60-r)/4} \cdot 5^{r/8}$$, where $$r = 0, 1, 2, \ldots, 60$$.

For a term to be rational, both exponents must be integers. The exponent of 3 is $$(60-r)/4$$, which is an integer when $$r \equiv 0 \pmod{4}$$. The exponent of 5 is $$r/8$$, which is an integer when $$r \equiv 0 \pmod{8}$$.

Both conditions are satisfied simultaneously when $$r$$ is divisible by $$\text{lcm}(4,8) = 8$$. So $$r \in \{0, 8, 16, 24, 32, 40, 48, 56\}$$, giving us 8 rational terms.

The total number of terms in the expansion is $$60 + 1 = 61$$. Therefore, the number of irrational terms is $$n = 61 - 8 = 53$$.

We need to check the divisibility of $$n - 1 = 52$$. We have $$52 = 4 \times 13 = 2^2 \times 13$$. Checking each option: $$52 / 26 = 2$$ (divisible), $$52 / 30$$ is not an integer, $$52 / 8 = 6.5$$ (not divisible), and $$52 / 7$$ is not an integer. Hence $$(n-1)$$ is divisible by 26.

We have to compare the coefficient of $$x^{7}$$ in the expansion $$\left(x^{2}+\dfrac{1}{bx}\right)^{11}$$ with the coefficient of $$x^{-7}$$ in the expansion $$\left(x-\dfrac{1}{bx^{2}}\right)^{11}$$, where $$b \neq 0$$.

First we consider $$\left(x^{2}+\dfrac{1}{bx}\right)^{11}$$. For a binomial of the form $$(A+B)^{n}$$, the general term is given by the Binomial Theorem: $$T_{r}=\binom{n}{r}A^{r}B^{\,n-r}.$$ Here, $$A=x^{2},\;B=\dfrac{1}{bx},\;n=11.$$

So the general term is $$T_{r}=\binom{11}{r}\left(x^{2}\right)^{r}\left(\dfrac{1}{bx}\right)^{11-r}.$$ Now we simplify the powers of $$x$$:

$$T_{r}=\binom{11}{r}x^{2r}\left(\dfrac{1}{b}\right)^{11-r}x^{-(11-r)} =\binom{11}{r}\left(\dfrac{1}{b}\right)^{11-r}x^{\,2r-(11-r)} =\binom{11}{r}\left(\dfrac{1}{b}\right)^{11-r}x^{\,3r-11}.$$

We need the exponent of $$x$$ to be $$7$$, so we set $$3r-11=7 \;\;\Longrightarrow\;\; 3r=18 \;\;\Longrightarrow\;\; r=6.$$

Substituting $$r=6$$ in the coefficient we obtain $$\text{Coefficient of }x^{7}= \binom{11}{6}\left(\dfrac{1}{b}\right)^{5}.$$ Since $$\binom{11}{6}=462$$, this coefficient becomes $$\dfrac{462}{b^{5}}.$$

Now we turn to $$\left(x-\dfrac{1}{bx^{2}}\right)^{11}$$. Again, using the Binomial Theorem with $$A=x,\;B=-\dfrac{1}{bx^{2}},\;n=11$$, the general term is

$$T_{r}=\binom{11}{r}x^{r}\left(-\dfrac{1}{bx^{2}}\right)^{11-r} =\binom{11}{r}x^{r}(-1)^{11-r}\left(\dfrac{1}{b}\right)^{11-r}x^{-2(11-r)}.$$

Combining the powers of $$x$$ we get $$T_{r}=\binom{11}{r}(-1)^{11-r}\left(\dfrac{1}{b}\right)^{11-r}x^{\,r-2(11-r)} =\binom{11}{r}(-1)^{11-r}\left(\dfrac{1}{b}\right)^{11-r}x^{\,r-22+2r} =\binom{11}{r}(-1)^{11-r}\left(\dfrac{1}{b}\right)^{11-r}x^{\,3r-22}.$$

We require the exponent of $$x$$ to be $$-7$$, hence $$3r-22=-7 \;\;\Longrightarrow\;\; 3r=15 \;\;\Longrightarrow\;\; r=5.$$

Substituting $$r=5$$ in the coefficient gives $$\text{Coefficient of }x^{-7}= \binom{11}{5}(-1)^{6}\left(\dfrac{1}{b}\right)^{6}.$$ Because $$(-1)^{6}=1$$ and $$\binom{11}{5}=462$$, this coefficient equals $$\dfrac{462}{b^{6}}.$$

The problem states that these two coefficients are equal, so we set

$$\dfrac{462}{b^{5}}=\dfrac{462}{b^{6}}.$$

Cancelling the common factor $$462$$ and cross-multiplying, we obtain

$$b^{6}=b^{5}\;\;\Longrightarrow\;\; b^{6}-b^{5}=0\;\;\Longrightarrow\;\; b^{5}(b-1)=0.$$

Since $$b\neq 0$$, the factor $$b^{5}$$ cannot be zero, leaving

$$b-1=0\;\;\Longrightarrow\;\; b=1.$$

Hence, the correct answer is Option C.

We are asked to look at the expansion of $$\left(x\sin\alpha + a\dfrac{\cos\alpha}{x}\right)^{10}$$ and pick out that term in which the power of $$x$$ is zero, that is, the term independent of $$x$$.

First we recall the binomial-theorem formula $$\left(u+v\right)^{n}=\sum_{r=0}^{n}\binom{n}{r}u^{\,n-r}v^{\,r},$$ where the general term is $$T_{r+1}=\binom{n}{r}u^{\,n-r}v^{\,r}.$$

In the present question we have $$u=x\sin\alpha,\qquad v=a\dfrac{\cos\alpha}{x},\qquad n=10.$$ So the general term is

$$T_{r+1}= \binom{10}{r}\bigl(x\sin\alpha\bigr)^{10-r}\left(a\dfrac{\cos\alpha}{x}\right)^{r}.$$

We simplify this power by power:

$$\bigl(x\sin\alpha\bigr)^{10-r}=x^{\,10-r}(\sin\alpha)^{10-r},$$ $$\left(a\dfrac{\cos\alpha}{x}\right)^{r}=a^{\,r}(\cos\alpha)^{r}x^{-\,r}.$$

Multiplying the two parts gives

$$T_{r+1}=\binom{10}{r}a^{\,r}(\sin\alpha)^{10-r}(\cos\alpha)^{r}\;x^{(10-r)-r} =\binom{10}{r}a^{\,r}(\sin\alpha)^{10-r}(\cos\alpha)^{r}\;x^{10-2r}.$$

For independence of $$x$$ we need the exponent of $$x$$ to be zero, so we set $$10-2r=0\quad\Longrightarrow\quad r=5.$$

Substituting $$r=5$$ back into the expression, the constant term becomes

$$T_{\text{indep}}=\binom{10}{5}a^{5}(\sin\alpha)^{5}(\cos\alpha)^{5}.$$

We know that $$\binom{10}{5}=\dfrac{10!}{5!\,5!},$$ so numerically the term is

$$T_{\text{indep}}=\dfrac{10!}{(5!)^{2}}\;a^{5}(\sin\alpha\cos\alpha)^{5}.$$

The problem says that the greatest value (maximum possible value) of this constant term is exactly $$\dfrac{10!}{(5!)^{2}}.$$ Hence we must have

$$\dfrac{10!}{(5!)^{2}}\;|a|^{5}\;|\sin\alpha\cos\alpha|^{5}= \dfrac{10!}{(5!)^{2}}.$$

Dividing both sides by the common factor $$\dfrac{10!}{(5!)^{2}}$$ we arrive at

$$|a|^{5}\;|\sin\alpha\cos\alpha|^{5}=1.$$

Now we recall the standard trigonometric result $$\sin\alpha\cos\alpha=\dfrac{\sin2\alpha}{2},$$ whose maximum absolute value is $$\dfrac{1}{2}.$$ Therefore the greatest possible value of $$|\sin\alpha\cos\alpha|^{5}$$ is $$\left(\dfrac12\right)^{5}=\dfrac{1}{32}.$$

Putting this maximum into our equation gives

$$|a|^{5}\;\dfrac{1}{32}=1 \quad\Longrightarrow\quad |a|^{5}=32 \quad\Longrightarrow\quad |a|=2.$$

So $$a$$ can be either $$2$$ or $$-2$$, but among the listed choices only the positive value $$2$$ is present.

Hence, the correct answer is Option D.

We need to find $$\sum_{r=0}^{6} \binom{6}{r} \cdot \binom{6}{6-r}$$.

Since $$\binom{6}{6-r} = \binom{6}{r}$$ (by the symmetry property of binomial coefficients), we can rewrite this as $$\sum_{r=0}^{6} \binom{6}{r} \cdot \binom{6}{r} = \sum_{r=0}^{6} \binom{6}{r}^2$$.

Now we use the Vandermonde identity. The identity states that $$\sum_{r=0}^{n} \binom{n}{r} \cdot \binom{n}{n-r} = \binom{2n}{n}$$. This comes from comparing the coefficient of $$x^n$$ on both sides of $$(1+x)^n \cdot (1+x)^n = (1+x)^{2n}$$.

Here $$n = 6$$, so $$\sum_{r=0}^{6} \binom{6}{r} \cdot \binom{6}{6-r} = \binom{12}{6}$$.

Computing $$\binom{12}{6} = \frac{12!}{6! \cdot 6!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{665280}{720} = 924$$.

The answer is $$924$$, which is Option D.

1. Simplify the Terms Using Logarithmic Properties

Using the fundamental property of logarithms where $$b^{\log_b y} = y$$, we can simplify both terms within the given binomial expression:

For the first term:

$$a = 3^{\log_3 \sqrt{25^{x-1}+7}} = \sqrt{25^{x-1}+7}$$

For the second term:

$$b = 3^{\left(-\frac{1}{8}\right)\log_3(5^{x-1}+1)} = 3^{\log_3 (5^{x-1}+1)^{-\frac{1}{8}}} = (5^{x-1}+1)^{-\frac{1}{8}}$$

Thus, the given expression simplifies directly to the standard binomial form:

$$(a + b)^{10} = \left(\sqrt{25^{x-1}+7} + (5^{x-1}+1)^{-\frac{1}{8}}\right)^{10}$$

2. Write the Ninth Term Formula

The general term $$T_{r+1}$$ in a binomial expansion of $$(a+b)^n$$ is defined by:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

For the ninth term ($$T_9$$), we substitute $$r = 8$$ and $$n = 10$$ into the expression:

$$T_9 = \binom{10}{8} a^{10-8} b^8$$

$$T_9 = \binom{10}{8} a^2 b^8$$

3. Substitute the Simplified Terms

Substitute the expressions for $$a$$ and $$b$$ into the ninth term formula:

$$T_9 = \binom{10}{8} \left(\sqrt{25^{x-1}+7}\right)^2 \left((5^{x-1}+1)^{-\frac{1}{8}}\right)^8$$

Now, evaluate each part step-by-step:

$$\binom{10}{8} = \frac{10 \times 9}{2 \times 1} = 45$$

$$\left(\sqrt{25^{x-1}+7}\right)^2 = 25^{x-1}+7$$

$$\left((5^{x-1}+1)^{-\frac{1}{8}}\right)^8 = (5^{x-1}+1)^{-1} = \frac{1}{5^{x-1}+1}$$

Combine these components together:

$$T_9 = 45 \cdot \frac{25^{x-1}+7}{5^{x-1}+1}$$

4. Solve the Algebraic Equation

We are given that the ninth term is equal to $$180$$:

$$45 \cdot \frac{25^{x-1}+7}{5^{x-1}+1} = 180$$

Divide both sides by $$45$$:

$$\frac{25^{x-1}+7}{5^{x-1}+1} = 4$$

$$25^{x-1}+7 = 4(5^{x-1}+1)$$

To solve this easily, let us use a substitution variable by setting $$5^{x-1} = t$$. Since $$25^{x-1} = (5^2)^{x-1} = (5^{x-1})^2 = t^2$$, the equation transforms into a quadratic equation:

$$t^2 + 7 = 4(t + 1)$$

$$t^2 + 7 = 4t + 4$$

$$t^2 - 4t + 3 = 0$$

Factorize the quadratic equation:

$$(t-1)(t-3) = 0$$

This gives two possible values for $$t$$:

$$t = 1 \text{ or } t = 3$$

5. Find the Values of x

Now, substitute back $$t = 5^{x-1}$$ to find the corresponding values of $$x$$:

Case 1: When $$t = 1$$

$$5^{x-1} = 1$$

$$5^{x-1} = 5^0$$

$$x - 1 = 0 \implies x = 1$$

Case 2: When $$t = 3$$

$$5^{x-1} = 3$$

$$\log_5(5^{x-1}) = \log_5 3$$

$$x - 1 = \log_5 3$$

$$x = 1 + \log_5 3$$

$$x = \log_5 5 + \log_5 3$$

$$x = \log_5 15$$

We are asked to find the value of the summation $$\displaystyle\sum_{r=0}^{20} r^{2}\binom{20}{r}$$ where $$\binom{20}{r}$$ is the coefficient of $$x^{r}$$ in the binomial expansion of $$(1+x)^{20}$$. Our task is purely algebraic, so we shall begin by recalling two standard binomial identities that come directly from the Binomial Theorem.

First, the Binomial Theorem states that

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

Now, two very useful consequences of this theorem are obtained by differentiating and then manipulating these expansions.

Identity 1 (Sum of first powers):

$$\sum_{r=0}^{n} r\binom{n}{r} \;=\; n\,2^{\,n-1}.$$

Identity 2 (Sum of product of consecutive integers):

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

We will use both identities. Observe that every term $$r^{2}$$ can be split as

$$r^{2} \;=\; r(r-1)\;+\;r.$$

Hence, for every $$r$$,

$$ r^{2}\binom{20}{r} \;=\; \bigl[r(r-1)\bigr]\binom{20}{r} \;+\; r\binom{20}{r}. $$

Adding term-by-term from $$r=0$$ to $$r=20$$ we have

$$ \sum_{r=0}^{20} r^{2}\binom{20}{r} \;=\; \sum_{r=0}^{20} r(r-1)\binom{20}{r} \;+\; \sum_{r=0}^{20} r\binom{20}{r}. $$

We can now invoke the two identities with $$n=20$$.

Using Identity 2 with $$n=20$$ gives

$$\sum_{r=0}^{20} r(r-1)\binom{20}{r} \;=\; 20\,(20-1)\,2^{\,20-2} \;=\; 20\times19\times2^{18}. $$

Using Identity 1 with $$n=20$$ gives

$$\sum_{r=0}^{20} r\binom{20}{r} \;=\; 20\,2^{\,20-1} \;=\; 20\times2^{19}. $$

Let us place both expressions over the same power of two so we can add them conveniently. Notice that $$2^{19}=2\cdot2^{18}$$. Therefore

$$ 20\times2^{19} \;=\; 20\times2\cdot2^{18} \;=\; 40\times2^{18}. $$

Now we combine the two partial sums:

$$\bigl[20\times19\times2^{18}\bigr] \;+\; \bigl[40\times2^{18}\bigr] \;=\; \bigl[20\times19 + 40\bigr]\,2^{18}.$$

Carrying out the arithmetic inside the bracket,

$$20\times19 \;=\; 380,$$

and hence

$$380 + 40 \;=\; 420.$

Substituting this total back, we obtain

$$\sum_{r=0}^{20} r^{2}\binom{20}{r} \;=\; 420\times2^{18}.$$

Therefore the given summation equals $$420\times2^{18}$$.

Hence, the correct answer is Option A.

We need the fourth term in the expansion of $$\left(x + x^{\log_2 x}\right)^7$$.

Using the binomial theorem, the general term is $$T_{r+1} = \binom{7}{r} x^{7-r} \cdot \left(x^{\log_2 x}\right)^r = \binom{7}{r} x^{7-r+r\log_2 x}$$.

For the fourth term, $$r = 3$$: $$T_4 = \binom{7}{3} x^{4 + 3\log_2 x} = 35 \cdot x^{4 + 3\log_2 x}$$.

We are given $$T_4 = 4480$$, so $$35 \cdot x^{4 + 3\log_2 x} = 4480$$, which gives $$x^{4 + 3\log_2 x} = 128$$.

Let $$\log_2 x = t$$, so $$x = 2^t$$. Substituting: $$(2^t)^{4 + 3t} = 128 = 2^7$$.

This gives $$2^{t(4 + 3t)} = 2^7$$, so $$t(4 + 3t) = 7$$, i.e., $$3t^2 + 4t - 7 = 0$$.

Using the quadratic formula: $$t = \frac{-4 \pm \sqrt{16 + 84}}{6} = \frac{-4 \pm 10}{6}$$.

So $$t = 1$$ or $$t = -\frac{7}{3}$$.

Since $$x \in \mathbb{N}$$, we need $$x = 2^t$$ to be a natural number. For $$t = 1$$: $$x = 2^1 = 2$$. For $$t = -\frac{7}{3}$$: $$x = 2^{-7/3}$$, which is not a natural number.

Therefore $$x = 2$$, which matches Option A.

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 need to find the value of $$-\binom{15}{1} + 2\binom{15}{2} - 3\binom{15}{3} + \ldots - 15\binom{15}{15} + \binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \ldots + \binom{14}{11}$$.

Consider the first part: $$S_1 = \sum_{r=1}^{15} (-1)^r \cdot r \cdot \binom{15}{r}$$.

We know that $$r \cdot \binom{15}{r} = 15 \cdot \binom{14}{r-1}$$. So $$S_1 = 15 \sum_{r=1}^{15} (-1)^r \binom{14}{r-1} = 15 \sum_{j=0}^{14} (-1)^{j+1} \binom{14}{j} = -15 \sum_{j=0}^{14} (-1)^j \binom{14}{j}$$.

Now $$\sum_{j=0}^{14} (-1)^j \binom{14}{j} = (1-1)^{14} = 0$$. So $$S_1 = 0$$.

For the second part: $$S_2 = \binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \ldots + \binom{14}{11}$$.

The sum of all odd-indexed binomial coefficients of order 14 is $$\binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \binom{14}{7} + \binom{14}{9} + \binom{14}{11} + \binom{14}{13} = 2^{13}$$.

Our sum $$S_2$$ is missing $$\binom{14}{13} = 14$$. So $$S_2 = 2^{13} - 14$$.

The total value is $$S_1 + S_2 = 0 + 2^{13} - 14 = 2^{13} - 14$$.

Hence, the correct answer is Option D.

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 $$1$$.

General term in the expansion of

$$\left(x+\frac{a}{x^2}\right)^n$$

is

$$T_{r+1}=\binom nr x^{n-r}\left(\frac{a}{x^2}\right)^r$$

$$=\binom nr a^r x^{n-3r}$$

Hence coefficients of third, fourth and fifth terms are:

$$T_3:\binom n2 a^2$$

$$T_4:\binom n3 a^3$$

$$T_5:\binom n4 a^4$$

Given,

$$\binom n2 a^2:\binom n3 a^3:\binom n4 a^4=12:8:3$$

Taking ratios,

$$\frac{\binom n2 a^2}{\binom n3 a^3}=\frac{12}{8}=\frac32$$

$$\frac{\frac{n(n-1)}2}{\frac{n(n-1)(n-2)}6}\cdot\frac1a=\frac32$$

$$\frac{3}{n-2}\cdot\frac1a=\frac32$$

$$a(n-2)=2\quad\cdots(1)$$

Also,

$$\frac{\binom n3 a^3}{\binom n4 a^4}=\frac83$$

$$\frac{\frac{n(n-1)(n-2)}6}{\frac{n(n-1)(n-2)(n-3)}{24}}\cdot\frac1a=\frac83$$

$$\frac{4}{n-3}\cdot\frac1a=\frac83$$

$$a(n-3)=\frac32\quad\cdots(2)$$

From (1) and (2),

$$\frac2{n-2}=\frac{3/2}{n-3}$$

$$4(n-3)=3(n-2)$$

$$4n-12=3n-6$$

$$n=6$$

Then from

$$a(n-2)=2,$$

$$4a=2$$

$$a=\frac12$$

Now term independent of $$x$$ occurs when

$$n-3r=0$$

$$6-3r=0$$

$$r=2$$

Hence required term is

$$T_3=\binom62\left(\frac12\right)^2$$

$$=15\cdot\frac14$$

$$=\frac{15}{4}$$

Therefore, the term independent of $$x$$ is

$$\boxed{\frac{15}{4}}$$

The general term in the expansion of $$\left(4^{1/4} + 5^{1/6}\right)^{120}$$ is:

$$T_{r+1} = \binom{120}{r} \left(4^{1/4}\right)^{120-r} \left(5^{1/6}\right)^r = \binom{120}{r} \cdot 2^{(120-r)/2} \cdot 5^{r/6}$$

For this term to be rational, both $$2^{(120-r)/2}$$ and $$5^{r/6}$$ must be rational, which requires both $$(120-r)/2$$ and $$r/6$$ to be non-negative integers.

The condition $$r/6 \in \mathbb{Z}$$ means $$r$$ is divisible by 6. Since 120 is divisible by 2, the condition $$(120-r)/2 \in \mathbb{Z}$$ means $$r$$ is even, i.e., divisible by 2. Thus $$r$$ must be divisible by $$\text{lcm}(6, 2) = 6$$.

With $$0 \leq r \leq 120$$ and $$r \equiv 0 \pmod{6}$$, the values are $$r = 0, 6, 12, 18, \ldots, 120$$, giving $$\frac{120}{6} + 1 = 21$$ rational terms.

We are asked to find the remainder obtained when the quantity $$3 \times 7^{22} + 2 \times 10^{22} - 44$$ is divided by $$18$$. In modular-arithmetic language, this means that we must evaluate the given expression modulo $$18$$, i.e. find its value in $$\pmod{18}$$.

First, we note that for any integers $$a$$ and $$b$$, the basic property of congruences is

$$a \equiv b \pmod{m} \;\;\Longrightarrow\;\; a \cdot c \equiv b \cdot c \pmod{m},$$

and

$$a \equiv b \pmod{m},\; c \equiv d \pmod{m} \;\;\Longrightarrow\;\; a \pm c \equiv b \pm d \pmod{m}.$$

We shall therefore replace each large power by its much smaller remainder modulo $$18$$ and then combine the results.

Step 1: Evaluate $$7^{22} \pmod{18}.$$

We first look for a pattern in the powers of $$7$$ modulo $$18$$:

$$7^1 = 7 \equiv 7 \pmod{18}.$$

$$7^2 = 49 \equiv 49 - 2\!\times\!18 = 49 - 36 = 13 \pmod{18}.$$

Now multiply again by $$7$$:

$$7^3 \;=\; 7^2 \times 7 = 13 \times 7 = 91 \equiv 91 - 5\!\times\!18 = 91 - 90 = 1 \pmod{18}.$$

We have reached $$7^3 \equiv 1 \pmod{18}.$$ This is extremely useful because now every third power will repeat the remainder $$1$$. In other words,

$$7^{3k} \equiv 1 \pmod{18}\quad\text{for every integer }k.$$

We write $$22$$ as a multiple of $$3$$ plus the remainder:

$$22 = 21 + 1 = 3 \times 7 + 1.$$

So

$$7^{22} = 7^{21 + 1} = 7^{21}\,7^1 = (7^3)^7 \, 7.$$

Because $$7^3 \equiv 1$$, we replace it by $$1$$:

$$(7^3)^7 \, 7 \equiv 1^7 \, 7 = 7 \pmod{18}.$$

Thus we have

$$7^{22} \equiv 7 \pmod{18}.$$

Step 2: Evaluate $$10^{22} \pmod{18}.$$

We again look for a pattern, this time in the powers of $$10$$ modulo $$18$$:

$$10^1 = 10 \equiv 10 \pmod{18}.$$

Now square that value:

$$10^2 = 100 \equiv 100 - 5\!\times\!18 = 100 - 90 = 10 \pmod{18}.$$

Notice that the remainder has come back to $$10$$. Therefore a second multiplication by $$10$$ will do the same:

$$10^3 = 10^2 \times 10 \equiv 10 \times 10 = 100 \equiv 10 \pmod{18}.$$

Hence once we reach the power $$2$$, every additional power leaves the remainder unchanged at $$10$$. Concretely, for all positive integers $$n \ge 1,$$

$$10^n \equiv 10 \pmod{18}.$$

In particular,

$$10^{22} \equiv 10 \pmod{18}.$$

Step 3: Substitute the reduced remainders into the original expression.

We have obtained

$$7^{22} \equiv 7 \pmod{18},$$

and

$$10^{22} \equiv 10 \pmod{18}.$$

Thus

$$3 \times 7^{22} + 2 \times 10^{22} - 44$$

becomes

$$3 \times 7 + 2 \times 10 - 44 \pmod{18}.$$

Carrying out the multiplications gives

$$21 + 20 - 44 \pmod{18}.$$

Add the first two numbers:

$$21 + 20 = 41,$$

so we now have

$$41 - 44 = -3 \pmod{18}.$$

Step 4: Convert the negative remainder to a positive one.

A remainder of $$-3$$ can be made positive by adding $$18$$ (because adding or subtracting any multiple of the modulus does not change a congruence):

$$-3 + 18 = 15.$$

Therefore

$$-3 \equiv 15 \pmod{18}.$$

The required non-negative remainder is thus $$15$$.

So, the answer is $$15$$.

For the binomial

the general term (using the Binomial Theorem $$\bigl(a+b\bigr)^n=\displaystyle\sum_{r=0}^{n}{^nC_r}\,a^{\,n-r}\,b^{\,r}$$) is

First we collect the powers of $$x$$ present in this term.

From the first factor we get $$x^{\,12-r}$$ and from the second factor $$x^{-2r}$$. Therefore the combined power of $$x$$ is

To obtain the term independent of $$x$$ we must set the exponent of $$x$$ equal to $$0$$:

Now we substitute $$r=4$$ back into the expression for the general term.

Simplifying step by step:

Because $$12=3\cdot4$$, we rewrite $$12^{4}=(3\cdot4)^{4}=3^{4}\,4^{4}$$. Substituting this gives

Now we put actual numerical values.

Thus

Observe that

Hence we can write

The problem states that the term independent of $$x$$ equals

(There is a typographical mix-up of the symbol in the statement, but comparing the obtained form with the required one, the numerical multiplier is clearly 55.)

Therefore,

So, the answer is $$55$$.

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 $$98$$.

We first simplify the expression inside the brackets. For the first fraction, note that $$x + 1 = (x^{1/3})^3 + 1 = (x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)$$, so $$\dfrac{x+1}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$$.

For the second fraction, factor the denominator: $$x - x^{1/2} = x^{1/2}(x^{1/2} - 1)$$, and the numerator: $$x - 1 = (x^{1/2} - 1)(x^{1/2} + 1)$$. So $$\dfrac{x-1}{x - x^{1/2}} = \dfrac{(x^{1/2}-1)(x^{1/2}+1)}{x^{1/2}(x^{1/2}-1)} = \dfrac{x^{1/2}+1}{x^{1/2}} = 1 + x^{-1/2}$$.

Subtracting: $$(x^{1/3} + 1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$$. We need the term independent of $$x$$ in $$(x^{1/3} - x^{-1/2})^{10}$$.

The general term is $$\binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r = \binom{10}{r}(-1)^r\,x^{(10-r)/3 - r/2}$$. Setting the exponent to zero: $$\dfrac{10-r}{3} - \dfrac{r}{2} = 0$$, which gives $$2(10-r) = 3r$$, so $$r = 4$$.

The term independent of $$x$$ is $$\binom{10}{4}(-1)^4 = 210$$.

We consider the binomial expansion formula first. For any real numbers $$a$$ and $$b$$ and a positive integer $$n$$, the expansion of $$(a+b)^n$$ is written as

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

Here, the general term (also called the $$(r+1)^\text{th}$$ term) is

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

Now, in our problem we have $$a = 2$$ and $$b = \dfrac{x}{3}$$, so the expansion of $$\left(2+\dfrac{x}{3}\right)^n$$ is

$$ \left(2+\dfrac{x}{3}\right)^n = \sum_{r=0}^{n} \binom{n}{r}\,2^{\,n-r}\left(\dfrac{x}{3}\right)^{r}. $$

From this, the coefficient of the term containing $$x^{r}$$ is obtained by looking at the power of $$x$$ in each term. Since $$\left(\dfrac{x}{3}\right)^r = \dfrac{x^{r}}{3^{\,r}},$$ the coefficient of $$x^{r}$$ is

$$ \binom{n}{r}\,2^{\,n-r}\,\dfrac{1}{3^{\,r}}. $$

We are told that the coefficients of $$x^{7}$$ and $$x^{8}$$ are equal. Writing these coefficients explicitly:

Coefficient of $$x^{7}$$

$$ C_7 = \binom{n}{7}\,2^{\,n-7}\,\dfrac{1}{3^{\,7}}. $$

Coefficient of $$x^{8}$$

$$ C_8 = \binom{n}{8}\,2^{\,n-8}\,\dfrac{1}{3^{\,8}}. $$

The condition given in the problem is

$$ C_7 = C_8. $$

Substituting the explicit expressions, we have

$$ \binom{n}{7}\,2^{\,n-7}\,\dfrac{1}{3^{\,7}} \;=\; \binom{n}{8}\,2^{\,n-8}\,\dfrac{1}{3^{\,8}}. $$

We start simplifying this equality step by step. First, divide both sides by the common positive factor $$2^{\,n-8}\,\dfrac{1}{3^{\,7}}$$ (which is definitely non-zero):

$$ \binom{n}{7}\;2^{\,n-7}\;\dfrac{1}{3^{\,7}} \;\Bigg/\; \Bigl(2^{\,n-8}\,\dfrac{1}{3^{\,7}}\Bigr) \;=\; \binom{n}{8}\;2^{\,n-8}\;\dfrac{1}{3^{\,8}} \;\Bigg/\; \Bigl(2^{\,n-8}\,\dfrac{1}{3^{\,7}}\Bigr). $$

Simplifying the left‐hand side first:

$$ \frac{2^{\,n-7}}{2^{\,n-8}} = 2^{\,1} = 2, \qquad \frac{1/3^{\,7}}{1/3^{\,7}} = 1. $$

So the left‐hand side simplifies to

$$ 2\,\binom{n}{7}. $$

Simplifying the right‐hand side likewise, observe

$$ \frac{1/3^{\,8}}{1/3^{\,7}} = \frac{1}{3}, \qquad \frac{2^{\,n-8}}{2^{\,n-8}} = 1, $$

so the right‐hand side becomes

$$ \frac{1}{3}\,\binom{n}{8}. $$

Equating these simplified expressions gives

$$ 2\,\binom{n}{7} \;=\; \frac{1}{3}\,\binom{n}{8}. $$

Multiplying both sides by $$3$$ to clear the denominator, we get

$$ 6\,\binom{n}{7} = \binom{n}{8}. $$

Next, we use the well-known relation between consecutive binomial coefficients, namely

$$ \frac{\binom{n}{7}}{\binom{n}{8}} = \frac{8}{n-7}. $$

This formula itself comes from expanding the factorial form:

$$ \frac{\binom{n}{7}}{\binom{n}{8}} = \frac{\displaystyle\frac{n!}{7!\,(n-7)!}} {\displaystyle\frac{n!}{8!\,(n-8)!}} = \frac{8!\,(n-8)!}{7!\,(n-7)!} = \frac{8}{n-7}. $$

Returning to our equation $$6\,\binom{n}{7} = \binom{n}{8},$$ divide both sides by $$\binom{n}{8}$$ (which is certainly non-zero for $$n \ge 8$$):

$$ 6\,\frac{\binom{n}{7}}{\binom{n}{8}} = 1. $$

Substituting the ratio $$\dfrac{\binom{n}{7}}{\binom{n}{8}} = \dfrac{8}{n-7}$$ gives

$$ 6 \times \frac{8}{n-7} = 1. $$

Hence