Question

The sum of series ${\cot }^{-1}\left(\frac{9}{2}\right)+{\cot }^{-1}\left(\frac{33}{4}\right)+{\cot }^{-1}\left(\frac{129}{8}\right)+\dots \dots \dots ...\infty$ is equal to

• A. ${\cot }^{-1}(2)$
• B. ${\cot }^{-1}3$
• C. ${\cot }^{-1}(-1)$
• D. ${\cot }^{-1}(1)$

Answer 1

Answer: (A)

$S={\cot }^{-1}\left(\frac{9}{2}\right)+{\cot }^{-1}\left(\frac{33}{4}\right)+{\cot }^{-1}\left(\frac{129}{8}\right)+\dots \infty$
$T_1={\cot }^{-1}\left(\frac{9}{2}\right)={\tan }^{-1}\left(\frac{2}{9}\right)={\tan }^{-1}\left(\frac{2}{1+4\times 2}\right)={\tan }^{-1}4-{\tan }^{-1}2$
$T_2={\cot }^{-1}\left(\frac{9}{2}\right)={\tan }^{-1}\left(\frac{4}{33}\right)={\tan }^{-1}\left(\frac{4}{1+8\times 4}\right)={\tan }^{-1}8-{\tan }^{-1}4$
$T_n={\tan }^{-1}{2}^{n+1}-{\tan }^{-1}{2}^n$
$S_n=\Sigma T_n={\tan }^{-1}{2}^{n+1}-{\tan }^{-1}2$
$\text{ as }n\to \infty$
$S_{\infty }=\frac{\pi }{2}-{\tan }^{-1}2={\cot }^{-1}2$