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..