Find $\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms Find $S=\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms
I first multiplied and divided $S$ with $1\cdot3\cdot5$
$$\frac{S}{15}=\frac{1}{1\cdot3\cdot5\cdot7}+\frac{1\cdot3}{1\cdot3\cdot5\cdot7\cdot9}+\frac{1\cdot3\cdot5}{1\cdot3\cdot5\cdot7\cdot9\cdot11}+\cdots$$
Using the expansion of $(2n)!$
$$1\cdot3\cdot5\cdots(2n-1)=\frac{(2n)!}{2^nn!}$$
$$S=15\left[\sum_{r=1}^{20}\frac{\frac{(2r)!}{2^rr!}}{\frac{(2(r+3))!}{2^{r+3}(r+3)!}}\right]$$
$$S=15\cdot8\cdot\left[\sum_{r=1}^{20}\frac{(2r)!}{r!}\cdot\frac{(r+3)!}{(2r+6)!}\right]$$
$$S=15\sum_{r=1}^{20}\frac{1}{(2r+5)(2r+3)(2r+1)}$$
How can I solve the above expression? Or is there an simpler/faster method?
 A: Hint: 
$\frac{1}{(2r+5)(2r+3)(2r+1)}=\frac{1}{4}\left(\frac{1}{(2r+3)(2r+1)}-\frac{1}{(2r+5)(2r+3)}\right)$
A: Here's another approach, using recurrence relations.
Note that
$$\begin{align}
a_1&=\frac 17\qquad\text{and}\\
a_n&=a_{n-1}\left(\frac{2n-1}{2n+5}\right)\\
(2n+1)a_n+4a_n&=(2n-1)a_{n-1}\\
a_n&=\frac 14\big(b_{n-1}-b_n\big)\\
\end{align}$$
where $b_m=(2m+1)a_m$.  
Summing from $2$ to $n$ by telescoping and adding $a_1$:
$$\begin{align}
\sum_{i=1}^{n}a_i
&=\frac 14(b_1-b_n)+a_1\\
&=\frac 14\big(1-(2n+1)a_n\big)
\qquad\qquad\qquad\text{as $b_1=3a_1$ and $a_1=\frac 17$}\\
&=\frac 14\bigg(1-\color{green}{(2n+1)}\cdot\frac{1\cdot 3\cdot 5\cdot \color{blue}{7\cdot 9\cdots(2n-1)}}{\color{blue}{7\cdot 9\cdot 11\cdots (2n-1)}\color{green}{(2n+1)}(2n+3)(2n+5)}\bigg)\\
&=\frac 14\bigg(1-\frac{15}{(2n+3)(2n+5)}\bigg)\\
&=\frac{n(n+4)}{(2n+3)(2n+5)}
\end{align}$$
Putting $n=20$: 
$$\begin{align}
\sum_{i=1}^{20}a_i
&=\frac{20\cdot 24}{43\cdot 45}=\frac{32}{129}\qquad\blacksquare
\end{align}$$
A: Alpha finds the sum to be $\frac {32}{129}$, giving an answer to your question of $1$.  The way to do it by hand is telescoping the series.
