<s>[INST] For the series $S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2$...... Problem : 
For the series  $$S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2+\frac{1}{(1+3+5+7)}(1+2+3+4)^2+\cdots $$ Find the nth term of the series. 
We know that nth can term of the series can be find by using $T_n = S_n -S_{n-1}$ 
$$S_n =1+ \sum \frac{(\frac{n(n+1)}{2})^2}{(2n-1)^2}$$ 
$$\Rightarrow S_n =\frac{n^4+5n^2+2n^3-4n+1}{(2n-1)^2}$$ 
But I think this is wrong, please suggest how to proceed thanks..
 [/INST] Consider the series
\begin{align}
S_{n} = 1 + \frac{(1+2)^{2}}{1+3} + \frac{(1+2+3)^{2}}{1+3+5} + \cdots + \frac{(1+2+\cdots+n)^{2}}{1+3+\cdots+(2n-1)}.
\end{align}
This series is seen as
\begin{align}
S_{n} &= 1 + \frac{1}{2^2}\binom{3}{2}^{2}+ \frac{1}{3^{2}} \binom{4}{2}^{2}+ \cdots + \frac{1}{n^{2}} \binom{n+1}{2}^{2} \\
&= \sum_{r=1}^{n} \frac{1}{r^{2}} \binom{r+1}{2}^{2} \\
&= \frac{1}{4} \sum_{r=1}^{n} (r+1)^{2} = \frac{1}{4} \sum_{r=2}^{n+1} r^{2} \\
&= \frac{1}{4} \left[ -1 + \frac{(n+1)(n+2)(2n+3)}{6} \right] \\
&= \frac{n}{24} \left(2 n^{2} + 9 n + 13 \right)
\end{align}
Making use of this formula it is quickly seen that
\begin{align}
S_{1} &= 1 \\
S_{2} &= 1 + \frac{3^{2}}{2^{2}} = 1 + \frac{(1+2)^{2}}{(1+3)}
\end{align}
</s>