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Find the sum of the following series to n terms $$\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$$

My attempt:

$$T_{n}=\frac{n^2}{(2n-1)(2n+1)}$$

I am unable to represent to proceed further. Though I am sure that there will be some method of difference available to express the equation. Please explain the steps and comment on the technique to be used with such questions.

Thanks in advance !

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Use partial fractions to get $$ \begin{align} \sum_{k=1}^n\frac{k^2}{(2k-1)(2k+1)} &=\sum_{k=1}^n\frac18\left(2+\frac1{2k-1}-\frac1{2k+1}\right)\\ &=\frac n4+\frac18-\frac1{16n+8}\\[3pt] &=\frac{(n+1)n}{4n+2} \end{align} $$ where we finished by summing a telescoping series.

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You can compute the partial fraction decomposition of $T_n$; $$T_n = \frac18 \left( \frac{1}{2n-1} - \frac{1}{2n+1} + 2 \right)$$

Then, you can separate the sum into three sum :

$$\sum_{n=1}^N T_n = \frac18 \left( \sum_{n=1}^N \frac{1}{2n-1} - \sum_{n=1}^N\frac{1}{2n+1} + \sum_{n=1}^N2 \right)$$ $$ = \frac18 \left( \sum_{n=2}^N \frac{1}{2n-1} - \sum_{n=1}^{N-1}\frac{1}{2n+1} + \sum_{n=1}^N2 + 1 - \frac{1}{2N+1}\right)$$ $$ = \frac18 \left( \sum_{n=1}^N2 + 1 - \frac{1}{2N+1}\right)$$ $$ = \frac18 \left( 2N + 1 - \frac{1}{2N+1}\right)$$ $$ = \frac18 \left( \frac{4N^2 +4N }{2N+1}\right)$$ $$ = \frac{N^2 +N}{2(2N+1)}$$

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  • $\begingroup$ can you explain the term $$1 - \frac{1}{2N+1}$$ in the 2nd step. How did you get that ? $\endgroup$ – cabmetric Sep 9 '15 at 15:31
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    $\begingroup$ Sure. I removed the term for $n=1$ in the first sum (it's the 1), and the term for $n=N$ of the second sum (the $-\frac1{2N+1}$), so that the two sums cancel each other. $\endgroup$ – Kevin Quirin Sep 9 '15 at 15:33
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Try to write $$T_n=A+\frac{B}{2n-1}+\frac{C}{2n+1}$$

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The $n$th term is $$n^2/(4n^2-1) =$$ $$ \frac{1}{4}.\frac {(4n^2-1)+1} {4n^2-1}=$$ $$\frac{1}{4} + \frac{1}{4}.\frac {1}{4n^2-1}=$$ $$\frac{1}{4}+ \frac {1}{4}. \left(\frac {1/2}{2n-1}- \frac {1/2}{2n+1}\right).$$ Is this enough?

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