Question

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

Answer 1

Let P (n) $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\cdots +\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$
For n = 1
$P\left(1\right)==\frac{1}{(2\times 1+1)(2\times 1+3)}=\frac{1}{3(2\times 1+3)}\Rightarrow \frac{1}{3\times 5}=\frac{1}{3\times 5}\Rightarrow \frac{1}{15}=\frac{1}{15}\therefore P\left(1\right)\text{is true}\text{Let P (n) be true for n = k}\therefore P\left(k\right)=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\cdots +\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)}\text{For n = k + 1}R.H.S.=\frac{k+1}{3(2k+5)}L.H.S.=\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)}=\frac{1}{(2k+3)}[frack3+\frac{1}{2k+5}]=\frac{1}{(2k+3)}\left[\frac{2k^2+5k+3}{3(2k+5)}\right]=\frac{1}{(2k+3)}\left[\frac{2k^2+2k+3k+3}{3(2k+5)}\right]=\frac{1}{(2k+3)}\left[\frac{(k+1)(2k+3)}{3(2k+5)}\right]=\frac{(k+1)}{3(2k+5)}$
$\therefore P(k+1)$ is true
Thus P(ki) is true $\Rightarrow$ p(k + 1) is true
by principle of mathematical induction,