Find the sum of the following series to n terms $\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$ 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 !
 A: 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 :
$$\sum_{n=1}^N T_n = \frac18 \left( \sum_{n=1}^N \frac{1}{2n-1} - \sum_{n=1}^N\frac{1}{2n+1} + \sum_{n=1}^N2 \right)$$
$$ = \frac18 \left( \sum_{n=2}^N \frac{1}{2n-1} - \sum_{n=1}^{N-1}\frac{1}{2n+1} + \sum_{n=1}^N2 + 1 - \frac{1}{2N+1}\right)$$
$$ = \frac18 \left( \sum_{n=1}^N2 + 1 - \frac{1}{2N+1}\right)$$
$$ = \frac18 \left( 2N + 1 - \frac{1}{2N+1}\right)$$
$$ = \frac18 \left( \frac{4N^2 +4N }{2N+1}\right)$$
$$ = \frac{N^2 +N}{2(2N+1)}$$
A: 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.
A: Try to write $$T_n=A+\frac{B}{2n-1}+\frac{C}{2n+1}$$
A: 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?
