Telescoping Series

$$S=\left(\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3^2}+\dfrac{1}{3}\times\dfrac{4}{7}+\dfrac{4^2}{7^2}\right)+\left(\dfrac{1}{3^3}+\dfrac{1}{3^2}\times\dfrac{4}{7}+\dfrac{1}{3}\times\dfrac{4^2}{7^2}+\dfrac{4^3}{7^3}\right)+......\infty$$

Take $$\dfrac{1}{3}=a$$ and $$\dfrac{4}{7}=b$$. 

$$S=\left(a+b\right)+\left(a^2+a\times b+b^2\right)+\left(a^3+a^2\times b+a\times b^2+b^3\right)+......\infty$$

Multiply both sides by (a-b). 

$$S\left(a-b\right)=\left(a+b\right)\left(a-b\right)+\left(a^2+a\times b+b^2\right)\left(a-b\right)+\left(a^3+a^2\times b+a\times b^2+b^3\right)\left(a-b\right)+......\infty$$

$$S\left(a-b\right)=\left(a^2-b^2\right)+\left(a^3-b^3\right)+\left(a^4-b^4\right)+......\infty$$

$$S\left(a-b\right)=\left(a^2+a^3+a^4+......\infty\right)-\left(b^2+b^3+b^4+........\infty\right)$$

We can see that $$\left(a^2+a^3+a^4+......\infty\right)$$ and $$\left(b^2+b^3+b^4+......\infty\right)$$ are in an infinite G.P.

$$\left(a^2+a^3+a^4+......\infty\right)=\dfrac{a^2}{1-a}$$

$$\left(b^2+b^3+b^4+......\infty\right)=\dfrac{b^2}{1-b}$$

$$S\left(a-b\right)=\left(\dfrac{a^2}{1-a}\right)-\left(\dfrac{b^2}{1-b}\right)$$

Now, insert the value of a and b in the above equation. 

$$S\left(\dfrac{1}{3}-\dfrac{4}{7}\right)=\left(\dfrac{\left(\dfrac{1}{3}\right)^2}{1-\dfrac{1}{3}}\right)-\left(\dfrac{\left(\dfrac{4}{7}\right)^2}{1-\dfrac{4}{7}}\right)$$

$$S\left(\dfrac{1}{3}-\dfrac{4}{7}\right)=\left(\dfrac{\dfrac{1}{9}}{\dfrac{2}{3}}\right)-\left(\dfrac{\dfrac{16}{49}}{\dfrac{3}{7}}\right)$$

$$S\left(\dfrac{1}{3}-\dfrac{4}{7}\right)=\left(\dfrac{1}{6}\right)-\left(\dfrac{16}{21}\right)$$

$$S\left(\dfrac{7-12}{21}\right)=\dfrac{7-32}{42}$$

$$S\times\dfrac{-5}{21}=\dfrac{-25}{42}$$

$$S=\dfrac{5}{2}$$

Hence, the sum of the given expression is $$\dfrac{5}{2}$$.

$$\therefore\ $$ The required answer is C. 

To find the sum of the first 25 terms of the given series, we first rewrite the general term $$a_n$$ using partial fractions to create a telescoping series.

The general term is given as:

$$a_n = \frac{-2}{4n^2 - 16n + 15}$$

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Step 1: Factorise the denominator

The quadratic expression in the denominator can be factored by splitting the middle term:

$$4n^2 - 16n + 15 = 4n^2 - 6n - 10n + 15 = 2n(2n - 3) - 5(2n - 3) = (2n - 3)(2n - 5)$$

Substituting this back into the expression for $$a_n$$:

$$a_n = \frac{-2}{(2n - 3)(2n - 5)}$$

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Step 2: Express $$a_n$$ in partial fractions

We can rewrite the numerator -2 as the difference between the two factors in the denominator:

$$-2 = (2n - 5) - (2n - 3)$$

Substituting this into $$a_n$$:

$$a_n = \frac{(2n - 5) - (2n - 3)}{(2n - 3)(2n - 5)} = \frac{2n - 5}{(2n - 3)(2n - 5)} - \frac{2n - 3}{(2n - 3)(2n - 5)}$$

$$a_n = \frac{1}{2n - 3} - \frac{1}{2n - 5}$$

This gives the telescoping structure $$a_n = V_n - V_{n-1}$$ where $$V_n = \frac{1}{2n - 3}$$.

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Step 3: Expand the sum up to 25 terms

Now, we evaluate the individual terms of the series from n = 1 to n = 25:

$$a_1 = \frac{1}{-1} - \frac{1}{-3} = -1 + \frac{1}{3}$$

$$a_2 = \frac{1}{1} - \frac{1}{-1} = 1 + 1$$

$$a_3 = \frac{1}{3} - \frac{1}{1} = \frac{1}{3} - 1$$

$$a_4 = \frac{1}{5} - \frac{1}{3}$$

$$\vdots$$

$$a_{24} = \frac{1}{45} - \frac{1}{43}$$

$$a_{25} = \frac{1}{47} - \frac{1}{45}$$

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Step 4: Cancel the intermediate terms

Summing all the terms together:

$$\sum_{n=1}^{25} a_n = \left(-1 + \frac{1}{3}\right) + (1 + 1) + \left(\frac{1}{3} - 1\right) + \left(\frac{1}{5} - \frac{1}{3}\right) + \dots + \left(\frac{1}{47} - \frac{1}{45}\right)$$

Grouping the positive and negative terms reveals the cancellation pattern:

$$\sum_{n=1}^{25} a_n = \left(\frac{1}{3} + 1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{47}\right) - \left(1 - \frac{1}{3} - 1 + 1 + \frac{1}{3} + \dots + \frac{1}{45}\right)$$

Most of the intermediate fractions cancel out, leaving behind only the following terms:

$$\sum_{n=1}^{25} a_n = \frac{1}{3} + \frac{1}{47}$$

$$\sum_{n=1}^{25} a_n = \frac{47 + 3}{3 \times 47} = \frac{50}{141}$$

Therefore, the value of $$a_1 + a_2 + \ldots + a_{25}$$ is equal to $$\frac{50}{141}$$.

The general term of the given series is $$T_k=\dfrac{4k}{\,1+4k^{4}\,},\;k\ge 1$$. We need the sum of the first ten terms, $$S_{10}=\displaystyle\sum_{k=1}^{10} T_k$$.

Step 1 - Convert $$T_k$$ into a telescoping form.
We try to express $$T_k$$ as a difference of two simpler rational terms. Write two quadratic expressions

$$A_k=2k^{2}-2k+1,\qquad B_k=2k^{2}+2k+1$$

First compute their product:

$$A_kB_k=(2k^{2}-2k+1)(2k^{2}+2k+1)=4k^{4}+1$$

Now observe

$$\dfrac{1}{A_k}-\dfrac{1}{B_k}= \dfrac{B_k-A_k}{A_kB_k}= \dfrac{4k}{\,4k^{4}+1\,}.$$

Hence

$$T_k=\dfrac{4k}{\,1+4k^{4}\,}= \dfrac{1}{2k^{2}-2k+1}-\dfrac{1}{2k^{2}+2k+1}.$$

Step 2 - Show the series telescopes.
Notice that

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

so $$\dfrac{1}{2k^{2}+2k+1}= \dfrac{1}{2(k+1)^{2}-2(k+1)+1}.$$ Define

$$a_k=\dfrac{1}{2k^{2}-2k+1}.$$

Then $$T_k = a_k - a_{k+1}.$$

Step 3 - Sum the first ten terms.
Because $$T_k=a_k-a_{k+1},$$ the partial sum is

$$S_{10}= \sum_{k=1}^{10}(a_k-a_{k+1}) =\bigl(a_1-a_2\bigr)+\bigl(a_2-a_3\bigr)+\dots+\bigl(a_{10}-a_{11}\bigr).$$

All intermediate terms cancel, leaving

$$S_{10}=a_1-a_{11}.$$

Step 4 - Evaluate $$a_1$$ and $$a_{11}$$.

$$a_1=\dfrac{1}{2(1)^2-2(1)+1}= \dfrac{1}{2-2+1}=1.$$

$$a_{11}=\dfrac{1}{2(11)^2-2(11)+1}= \dfrac{1}{242-22+1}= \dfrac{1}{221}.$$

Step 5 - Compute $$S_{10}$$.

$$S_{10}=1-\dfrac{1}{221}= \dfrac{221-1}{221}= \dfrac{220}{221}.$$

The fraction $$\dfrac{220}{221}$$ is already in lowest terms because $$221=13\cdot17$$ shares no common factor with $$220=2^{2}\cdot5\cdot11$$.

Thus $$m=220,\;n=221\; \Longrightarrow\; m+n=220+221=441.$$

Hence the required value is $$\boxed{441}$$.