Question

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

• A. $10/21$
• B. $13/27$
• C. $15/22$
• D. $8/33$

Answer 1

Answer: (A)

$10/21$

Given,

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

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

$\frac{1}{2}[\frac{(2+1)-(2-1)}{(2+1)(2-1)}+\frac{(4+1)-(4-1)}{(4+1)(4-1)}+......$

$\frac{1}{2}[\frac{1}{2-1}-\frac{1}{2-1}+\frac{1}{4-1}-\frac{1}{4+1}+....$

$\frac{1}{2}[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+.....$

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

$=\frac{1}{2}\times \frac{20}{21}$

$=\frac{10}{21}$