Question

$co{t}^{-1}(2\times 1^2)+co{t}^{-1}(2\times 2^2)+co{t}^{-1}(2\times 3^2)+...$ up to $\infty$ is equal to

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

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{4}$

Explanation for the correct option:

Given, $co{t}^{-1}2{\left(1\right)}^2+co{t}^{-1}2{\left(2\right)}^2+co{t}^{-1}2{\left(3\right)}^2+co{t}^{-1}2{\left(4\right)}^2+...$

$\therefore T_n=co{t}^{-1}\left(2n^2\right)$

$={\tan }^{-1}\frac{1}{2n^2}$
$={\tan }^{-1}\left[\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right]$

$={\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-1)$ ; $\left[\mathbf{\because }\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{A-B}{1+AB}\right)\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathbf{-}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{B}\right]$

$\therefore T_1={\tan }^{-1}3-{\tan }^{-1}1$

$T_2={\tan }^{-1}5-{\tan }^{-1}3$

$T_3={\tan }^{-1}7-{\tan }^{-1}5$

$\therefore S_n={\tan }^{-1}(2n+1)-{\tan }^{-1}\left(1\right)$

$\therefore \underset{n\to \infty }{\lim }{S}_n={\tan }^{-1}\infty -\frac{\pi }{4}$

$=\frac{\pi }{2}-\frac{\pi }{4}$

$=\frac{\mathbf{\pi }\mathbf{}}{\mathbf{4}}$

Hence, Option ‘A’ is Correct.