Question

What is the value of $\frac{1}{2^2-1}+\frac{1}{4^2-1}+\frac{1}{6^2-1}+...........\frac{1}{{20}^2-1}?$

Answer 1

$(\frac{1}{2^2}-1)+(\frac{1}{4^2}-1)+(\frac{1}{6^2}-1)+.........(\frac{1}{{20}^2}-1)$

So, the general term is

$\Rightarrow S_n={\sum }_{r=1}^n\frac{1}{{\left(2r\right)}^2-1^2}$

$S_{10}={\sum }_{r=1}^{10}\frac{1}{(2r-1)(2r+1)}$

$S_{10}={\sum }_{r=1}^{10}\frac{1}{2}\times (\frac{1}{2r-1}-\frac{1}{2r+1})$

$S_{10}=\frac{1}{2}\times (1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}.....\frac{1}{19}-\frac{1}{21})$

This is a telescopic series where term camel off each other.

$\Rightarrow S_{10}=\frac{1}{2}(1-\frac{1}{21})$

$\Rightarrow S_{10}=\frac{10}{21}$

Hence, the answer is $\frac{10}{21}.$