Question

The value of ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+\cdots +{\tan }^{-1}\left(\frac{1}{n^2+n+1}\right)$ is

• A. ${\tan }^{-1}\left(\frac{n}{n+2}\right)$
• B. ${\tan }^{-1}\left(\frac{n+1}{n-1}\right)$
• C. ${\tan }^{-1}\left(\frac{n-1}{n+2}\right)$
• D. ${\tan }^{-1}\left(\frac{n+2}{n-2}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{n}{n+2}\right)$

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

Hence required sum is $=\sum T_n$

$=\left[{\tan }^{-1}\right(2)-{\tan }^{-1}1]+\left[{\tan }^{-1}\right(3)-{\tan }^{-1}2]+........+{\tan }^{-1}\left(n\right)-{\tan }^{-1}(n-1)]+[{\tan }^{-1}(n+1)-{\tan }^{-1}n]$

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