Prove the sum to n terms of the series Prove that the sum to $n$ terms of a sequence:
$$\frac{1^2}{1\times 3}+\frac{2^2}{3\times 5}+\frac{3^2}{5\times 7}+\cdots $$
is 
$$
\frac{n(n+1)}{2(2n+1)}
$$
 A: If you don't notice the telescoping sum, since you already know the answer in advance you can take a more 'dumbass' approach.
Another way of writing this is $$\sum_{i=1}^n\frac{i^2}{(2i-1)(2i+1)}=\frac{n(n+1)}{2(2n+1)}$$
We can prove this is true by induction. The formula holds when $n=1$, because
$$\frac{1^2}{1\times3}=\frac{1(1+1)}{2(2\cdot1+1)}$$
Suppose the formula holds when $n=k$, for some positive integer $k$. Then:
\begin{align*}
\sum_{i=1}^{k+1}\frac{i^2}{(2i-1)(2i+1)} &= \left(\sum_{i=1}^k\frac{i^2}{(2i-1)(2i+1)}\right)+\frac{(k+1)^2}{(2(k+1)-1)(2(k+1)+1)} \\
&= \frac{k(k+1)}{2(2k+1)}+\frac{(k+1)^2}{(2k+1)(2k+3)} \\
&= \frac{k+1}{2k+1}\left(\frac{k}{2}+\frac{k+1}{2k+3}\right) \\
&= \frac{k+1}{2k+1}\left(\frac{k(2k+3)+2(k+1)}{2(2k+3)}\right) \\
&= \frac{k+1}{2k+1}\left(\frac{(2k+1)(k+2)}{2(2k+3)}\right) \\
&= \frac{(k+1)((k+1)+1)}{2(2(k+1)+1)}
\end{align*}
Which is exactly what the formula should be when $n=k+1$, so by induction the formula holds for all $n\in\mathbb{N}$.
A: Assuming that $n^2, (2n-1), (2n+1)$ are the general terms for $\{1^2, 2^2, 3^3, \cdots, \}$, $\{1, 3, 5,  \cdots, \}$, $\{3, 5, 7 \cdots, \}$ we see that $t_n=\frac{n^2}{(2n-1)(2n+1)}$ which partial fraction decomposition gives as $\frac 18[\frac{1}{2n-1}-\frac{1}{2n+1}]$. take the sum over $n$ from 1 to $n$ and get desired answer.
