Question

$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+..............+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

Answer 1

$\left(\frac{1}{3\times 5}\right)=\left(\right(\frac{1}{3})-(\frac{1}{5}\left)\right)\times \left(\frac{1}{2}\right)\left(\frac{1}{5\times 7}\right)=\left(\right(\frac{1}{5})-(\frac{1}{7}\left)\right)\times \left(\frac{1}{2}\right)\therefore sum=\left(\frac{1}{2}\right)\left[\right(\left(\frac{1}{3}\right)-\left(\frac{1}{5}\right)+\left(\frac{1}{5}\right)-\left(\frac{1}{7}\right)+......\left(\right(\frac{1}{2n+1})-(\frac{1}{2n+3}\left)\right)]=(\frac{1}{2}\left)\right[\left(\frac{1}{3}\right)-\left(\frac{1}{2n+3}\right)]=(\frac{1}{2}\left)\right(\frac{2n+3-3}{3(2n+3)})=(\frac{n}{3(2n+3)})$