Question

The sum of the infinite series
${\cot }^{-1}(2\cdot 1^2)+{\cot }^{-1}(2\cdot 2^2)+{\cot }^{-1}(2\cdot 3^2)+{\cot }^{-1}(2\cdot 4^2)+.......$ is equal to

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

Answer 1

Answer: (B)

$S=co{t}^{-1}2\cdot 1^2+co{t}^{-1}2\cdot 2^2$
$+co{t}^{-1}2\cdot 3^2+co{t}^{-1}2\cdot 4^2+.....$
$T_n=co{t}^{-1}\left(2\cdot n^2\right)$
$=ta{n}^{-1}\frac{1}{2n^2}=ta{n}^{-1}\left(\frac{2}{4n^2}\right)$
$=ta{n}^{-1}\left[\frac{\left(2n+1\right)-\left(2n-1\right)}{1+\left(4n^2-1\right)}\right]$
$=ta{n}^{-1}\left(2n+1\right)-ta{n}^{-1}\left(2n-1\right)$
$\therefore S=ta{n}^{-1}3-ta{n}^{-1}1+ta{n}^{-1}5-ta{n}^{-1}3$
$+ta{n}^{-1}7+ta{n}^{-1}5+.....\infty$
$=ta{n}^{-1}\infty -ta{n}^{-1}1$
$=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$