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..
 A: 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}
A: 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.
A: 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 $$T_n=\dfrac{(1+2+3+..+n)^2}{1+3+5+...+(2n-1)}=\dfrac{(n+1)^2}{4}$$ $$S_n=\sum_{k=1}^nT_k\\=\sum_{k=1}^n\dfrac{(k+1)^2}{4}\\=\sum_{k=1}^{n+1}\dfrac{k^2}{4}-\dfrac{1}{4}\\=\dfrac{(n+1)(n+2)(2n+3)}{24}-\dfrac{1}{4}$$
