Question

Let $f\left(n\right)=\sum _{k=1}^n{\cot }^{-1}(k^2+k+1).$ Then which of the following options is CORRECT ?

• A. $f\left(99\right)={\tan }^{-1}\left(100\right)-\frac{\pi }{2}$
• B. $f\left(99\right)={\tan }^{-1}\left(100\right)+\frac{\pi }{4}$
• C. $f\left(100\right)={\tan }^{-1}\left(100\right)-\frac{\pi }{4}$
• D. $f\left(100\right)={\tan }^{-1}\left(101\right)-\frac{\pi }{4}$

Answer 1

The correct option is D $f\left(100\right)={\tan }^{-1}\left(101\right)-\frac{\pi }{4}$

$f\left(n\right)=\sum _{k=1}^n{\cot }^{-1}(k^2+k+1)$

Let $f\left(k\right)={\cot }^{-1}(k^2+k+1)$
$={\tan }^{-1}\left(\frac{1}{k^2+k+1}\right)$
$={\tan }^{-1}\left[\frac{(k+1)-k}{1+(k+1)k}\right]={\tan }^{-1}(k+1)-{\tan }^{-1}\left(k\right)$

Thus, the required sum of $n$ terms of the given series is
$f\left(n\right)=({\tan }^{-1}2-{\tan }^{-1}1)+({\tan }^{-1}3-{\tan }^{-1}2)...+\left({\tan }^{-1}\right(n+1)-{\tan }^{-1}n)$

$\Rightarrow f\left(n\right)={\tan }^{-1}(n+1)-{\tan }^{-1}1$
$\Rightarrow f\left(n\right)={\tan }^{-1}(n+1)-\frac{\pi }{4}$