Question

$\cot^{-1} (2.1^{2})+\cot^{-1} (2.2^{2})+\cot^{-1}(2.3^2)+$.........up to $\infty$=

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

Answer 1

Answer: (B)

$\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots \infty$
$=\sum\limits_{r=1}^{\infty} \cot ^{-1}\left(2 \cdot r^{2}\right)=\sum\limits_{r=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 r^{2}}\right)$
$=\sum\limits_{r=1}^{\infty} \tan ^{-1}\left(\frac{(1+2 r)+(1-2 r)}{1-(1+2 r)(1-2 r)}\right)$
$=\sum\limits_{r=1}^{\infty}\left[\tan ^{-1}(1+2 r)+\tan ^{-1}(1-2 r)\right]$
$=\tan ^{-1} 3-\tan ^{-1} 1+\tan ^{-1} 5-\tan ^{-1} 3$
$+\tan ^{-1} 7-\tan ^{-1} 5+\ldots+\tan ^{-1} \infty$
$=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}$