Question:
If $\sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^{2}}=p$, then the value of $\tan p$ is :
Correct Option: , 2
Solution:
$\sum_{r=1}^{50} \tan ^{-1}\left(\frac{2}{4 r^{2}}\right)=\sum_{r=1}^{50} \tan ^{-1}\left(\frac{(2 r+1)-(2 r-1)}{1+(2 r+1)(2 r-1)}\right)$
$\sum_{r=1}^{50} \tan ^{-1}(2 r+1)-\tan ^{-1}(2 r-1)$
$\tan ^{-1}(101)-\tan ^{-1} 1 \Rightarrow \tan ^{-1} \frac{50}{51}$