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..

First we should note that $$1+2+3+...+n=\dfrac{n(n+1)}{2}$$ and $$1+3+5+...+(2n-1)=n^2.$$ Therefore general term in your series become to $$T_n=\dfrac{(1+2+3+..+n)^2}{1+3+5+...+(2n-1)}=\dfrac{(n+1)^2}{4}$$ $$S_n=\sum_{k=1}^nT_k\\=\sum_{k=1}^n\dfrac{(k+1)^2}{4}\\=\sum_{k=1}^{n+1}\dfrac{k^2}{4}-\dfrac{1}{4}\\=\dfrac{(n+1)(n+2)(2n+3)}{24}-\dfrac{1}{4}$$
