sequences and series - Find $frac{1}{7}+frac{1cdot3}{7cdot9}+frac{1cdot3cdot5}{7cdot9cdot11}+cdots$ 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?

Answer

Hint:

$\frac{1}{(2r+5)(2r+3)(2r+1)}=\frac{1}{4}\left(\frac{1}{(2r+3)(2r+1)}-\frac{1}{(2r+5)(2r+3)}\right)$