CAT 2000 — QA Question 4

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

$=\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+...+\frac{1}{19\times 21}$

$=\frac{1}{2}\left[\left(1-\frac{1}{3}\right]+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)...+\left(\frac{1}{19}-\frac{1}{21}\right)\right)$

$=\frac{1}{2}\left[\left(1-\frac{1}{21}\right]\right)=\frac{1}{2}\left[\frac{20}{21}\right]=\frac{10}{21}$

Hence, option (c).