Question

If $\sum _{r=1}^{50}{\tan }^{-1}\frac{1}{2r^2}=p,$ then the value of $\tan p$ is

• A. $100$
• B. $\frac{50}{51}$
• C. $\frac{101}{102}$
• D. $\frac{51}{50}$

Answer 1

The correct option is B $\frac{50}{51}$

$\because {\tan }^{-1}\frac{1}{2r^2}={\tan }^{-1}\left(\frac{2}{1+(4r^2-1)}\right)$
$={\tan }^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right)$
$={\tan }^{-1}(2r+1)-{\tan }^{-1}(2r-1)$
$\therefore \sum _{r=1}^{50}{\tan }^{-1}\left(\frac{1}{2r^2}\right)=\sum _{r=1}^{50}\left({\tan }^{-1}\right(2r+1)-{\tan }^{-1}(2r-1\left)\right)$
$\therefore p={\tan }^{-1}\left(101\right)-{\tan }^{-1}1$
$={\tan }^{-1}\left(\frac{101-1}{1+101\cdot 1}\right)$
$\therefore \tan p=\frac{50}{51}$