Question

$\frac{1}{3^2-1}+\frac{1}{5^2-1}+\frac{1}{7^2-1}+\cdots +\frac{1}{(201{)}^2-1}$ is equal to:

• A. $\frac{101}{404}$
• B. $\frac{101}{408}$
• C. $\frac{99}{400}$
• D. $\frac{25}{101}$

Answer 1

The correct option is D $\frac{25}{101}$

$S=\sum _{r=1}^{100}\frac{1}{(2r+1{)}^2-1}=\sum _{r=1}^{100}\frac{1}{(2r+2)\cdot 2\left(r\right)}$
$\therefore S=\frac{1}{4}\sum _{r=1}^{100}|\frac{1}{r}-\frac{1}{r+1}|$
$S=\frac{1}{4}((1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{100}-\frac{1}{101}))$
$\therefore S=\frac{1}{4}\left|\frac{100}{101}\right|=\frac{25}{101}$