Question

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

Answer 1

$P\left(n\right)-\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}.............\frac{1}{(2n+1)(2n+3)}$

$m=1$

$LHS=\frac{1}{15}$ $RHS=\frac{1}{3(2+3)}=\frac{1}{15}$

$P\left(n\right)$ is true for $n=1$

Assume $P\left(k\right)$ is true

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

$LHS$

$\frac{1}{3.5}+\frac{1}{5.7}...........\frac{1}{\left(2\right(k+1)+1)\left(2\right(k+1)+3)}$

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

$=\frac{1}{3.5}+\frac{1}{5.7}..........\frac{1}{(2k+3)(2k+5)}$

$RHS$

$\frac{(k+1)}{3\left(2\right(k+1)+3)}$

$=\frac{k+1}{3(2k+5)}$

$=\frac{k+1}{3(2k+5)}$

$=\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{(2k+1)(2k+3)}+\frac{1}{(2k+3)(2k+5)}$

$=\frac{K}{3(2k+3)}+\frac{(2k+3)(2k+5)}{\Rightarrow }\frac{1}{(2k+3)}[\frac{K}{3}+\frac{1}{(2k+5)}]$

$=\frac{1}{(2k+3)}\left(\frac{2k^2+5k+3}{3(2k+5)}\right)$

$=\frac{1}{(2k+3)}\left[\frac{2k^2+2k+3k+3}{3(2k+5)}\right]\Rightarrow \frac{1}{2k+3}\left[\frac{(2k+3)(k+1)}{3(2k+5)}\right]$

$=\frac{k+1}{3(2k+5)}=RHS$