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

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

Put n = 1

$\frac{1}{3.5}=\frac{1}{3\left(5\right)}$

$\frac{1}{15}=\frac{1}{15}$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

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

We have to show that,

$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.......+\frac{1}{(2k+1)(2k+3)}+\frac{1}{(2k+3)(2k+5)}=\frac{(k+1)}{3(2k+5)}$

Now,

$\{\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{(2k+1)(2k+3)}\}+\frac{1}{(2k+3)(2k+5)}$

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

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

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

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

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

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

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

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

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N by PMI.