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..

You don't need to bother with calculating $S_n - S_{n-1}$ here as the terms are explicitly given in the summation.
Here $T_n = \frac{(1+2+...+n)^2}{(1+3+5+...+(2n-1))}$
The numerator is the square of the first $n$ integers, so is equal to $(\frac{1}{2}n(n+1))^2$.
The denominator is the sum of the first $n$ odd integers. You can calculate the sum in a couple of ways, by direct application of the arithmetic series sum formula, or by subtracting the sum of even numbers from the sum of the first $2n$ integers. Either way, you'll figure out the denominator is equal to $n^2$. Now just do the division.
