Question

Find the sum of n terms of the series
$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..........$

Answer 1

$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+...........$

$n_r=\frac{1}{(2r-1)(2r+1)}$

$=\frac{2}{2}\frac{1}{(2r-1)(2r+1)}$

$=(2r+1)-(2r-1)2(2r-1)(2r+1)$

$=\frac{1}{2}[\frac{2r+1}{(2r-1)(2r+1)}-\frac{2r-1}{(2r-1)(2r+1)}]$

$=\frac{1}{2}[\frac{1}{2r-1}-\frac{1}{2r+1}]$

Put $r=1,u_1=\frac{1}{2}[\frac{1}{1}-\frac{1}{3}]$

$r=2,u_2=\frac{1}{2}[\frac{1}{3}-\frac{1}{5}]$

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$r=n,u_n=\frac{1}{2}[\frac{1}{2n-1}-\frac{1}{2n+1}]$

$S_n={\sum }_{r=1}^n{n}_r=\frac{1}{2}[\frac{1}{1}-\frac{1}{2n+1}]$

$=\frac{1}{2}\left[\frac{2n+1-1}{2n+1}\right]$

$S_n=\frac{n}{2n+1}$