Mathematics - Sequence & Series Question with Solution | TestHub

Given $S=\frac{1}{3^2-1}+\frac{1}{5^2-1}+\frac{1}{7^2-1}+\dots +\frac{1}{(201{)}^2-1}$

$T_n=\frac{1}{(2n+1{)}^2-1}=\frac{1}{4\left(n\right)(n+1)}$

$T_n=\frac{1}{4}\left(\frac{(n+1)-n}{n(n+1)}\right)=\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+1}\right)$

$\therefore S=\sum _{n=1}^{100}{T}_n=\sum _{n=1}^{100}\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+1}\right)$

$\therefore S=\frac{1}{4}\left(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{100}-\frac{1}{101}\right)\right)$

$\Rightarrow S=\frac{1}{4}\left(1-\frac{1}{101}\right)=\frac{1}{4}\left(\frac{100}{101}\right)=\frac{25}{101}$