Question

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

• A. $1.03$
• B. $1.00$
• C. $1.01$
• D. $1.02$

Answer 1

The correct option is C $1.01$

$R.H.S.=\sum _{n=1}^{100}{\cot }^{-1}\left(2n^2\right)=\sum _{n=1}^{100}{\tan }^{-1}\left(\frac{2}{4n^2}\right)=\sum _{n=1}^{100}{\tan }^{-1}\left(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right)=\sum _{n=1}^{100}{\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-1)={\tan }^{-1}\left(201\right)-{\tan }^{-1}\left(1\right)={\tan }^{-1}\left(\frac{200}{202}\right)={\cot }^{-1}\left(\frac{101}{100}\right)\Rightarrow {\cot }^{-1}\left(\alpha \right)={\cot }^{-1}\left(1.01\right)\Rightarrow \alpha =1.01$