If ${\cot }^{-1}\left(\alpha \right)={\cot }^{-1}2+{\cot }^{-1}8+{\cot }^{-1}18+{\cot }^{-1}32+\dots .$ upto $100$ terms, then $\alpha$ is:

Given ${\cot }^{-1}\left(\alpha \right)={\cot }^{-1}\left(2\right)+{\cot }^{-1}\left(8\right)+{\cot }^{-1}\left(18\right)+\dots .$

$=\sum _{n=1}^{100}{\tan }^{-1}\left(\frac{2}{4n^2}\right)$

$=\sum _{n=1}^{100}{\tan }^{-1}\left(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right)$

$=\sum _{n=1}^{100}{\tan }^{-1}\left(2n+1\right)-{\tan }^{-1}\left(2n-1\right)$

$={\tan }^{-1}201-{\tan }^{-1}1$

$={\tan }^{-1}\left(\frac{200}{202}\right)$

$\therefore {\cot }^{-1}\left(\alpha \right)={\cot }^{-1}\left(\frac{202}{200}\right)$

$\alpha =1.01$