Find the sum to the n terms of the series :
$\displaystyle \frac{1}{(2r-1)(2r+1)(2r+3)}$
Hello thereddevilsUsing Partial Fractions, you find that
$\displaystyle \frac{1}{(2r-1)(2r+1)(2r+3)} = \frac{1}{8}\left(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\right)$
So if the sum to n terms is $\displaystyle S$:
$\displaystyle 8S = \sum_{r=1}^n\left(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\right)$
$\displaystyle = \sum_{r=1}^n\frac{1}{2r-1} -\sum_{r=2}^{n+1}\frac{2}{2r-1}+\sum_{r=3}^{n+2}\frac{1}{2r-1}$
$\displaystyle = 1 + \frac{1}{3}+\sum_{r=3}^n\frac{1}{2r-1}-(\frac{2}{3}+2\sum_{r=3}^n\frac{1}{2r-1} + \frac{2}{2n+1})+\sum_{r=3}^n\frac{1}{2r-1}+\frac{1}{2n+1}+\frac{1}{2n+3}$
$\displaystyle =\frac{2}{3}-\frac{2}{2n+1}+\frac{1}{2n+1}+\frac{1}{2n+3}$
$\displaystyle \Rightarrow S=\frac{1}{8}\left(\frac{2}{3}-\frac{1}{2n+1}+\frac{1}{2n+3}\right)$
Grandad