Question

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

• A. $\frac{9}{19}$
• B. $\frac{10}{19}$
• C. $\frac{10}{21}$
• D. $\frac{11}{21}$

Answer 1

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

$n$ th term in the given series can be written as

$\frac{1}{(2n{)}^2-1}$

$=\frac{1}{(2n+1)(2n-1)}$

$=\frac{1}{2}\{\frac{1}{2n-1}-\frac{1}{2n+1}\}$

So, the given series can be written as

$\frac{1}{2}(1-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+..............\frac{1}{2}(\frac{1}{19}-\frac{1}{21})$

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

$=\frac{10}{21}$