Question

$\cot ^{-1}\left(2.1^{2}\right)+\cot ^{-1}\left(2.2^{2}\right)+\cot ^{-1}\left(2.3^{2}\right)+\ldots$ is equal to

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $\frac{\pi}{5}$

Answer 1

Answer: (A)

$\cot ^{-1}\left(2.1^{2}\right)+\cot ^{-1}\left(2.2^{2}\right)+\cot ^{-1}\left(2.3^{2}\right)+\ldots$
$=\displaystyle\sum_{r=1}^{\infty} \cot ^{-1}\left(2 \cdot r^{2}\right)=\displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 r^{2}}\right)$
$=\displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left[\frac{(1+2 r)+(1-2 r)}{1-(1+2 r)(1-2 r)}\right]$
$=\displaystyle\sum_{r=1}^{\infty}\left[\tan ^{-1}(1+2 r)+\tan ^{-1}(1-2 r)\right]$
$=\tan ^{-1} 3-\tan ^{-1} 1+\tan ^{-1} 5-\tan ^{-1} 3+\tan ^{-1} 7$
$-\tan ^{-1} 5+\ldots .+\tan ^{-1} \infty$
$=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}$