Question

The value fo the following expression is
$\left[\frac{1}{\left(2^2-1\right)}\right]+\left[\frac{1}{\left(4^2-1\right)}\right]+\left[\frac{1}{\left(6^2-1\right)}\right]+........+\left[\frac{1}{\left({20}^2-1\right)}\right]$ is

• A. $\frac{10}{21}$
• B. $\frac{13}{27}$
• C. $\frac{15}{22}$
• D. $\frac{8}{33}$

Answer 1

The correct option is A $\frac{10}{21}$

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

$=\frac{1}{2}\left[\frac{1}{3}\right]+\left[\frac{1}{15}\right]+\left[\frac{1}{35}\right]+.........\left[\frac{1}{399}\right]$.

$=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+......\frac{1}{2}(\frac{1}{19}-\frac{1}{21}).$

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

$=\frac{1}{2}[1-\frac{1}{21}]=\frac{10}{21}$

$A$ is correct.