Question

$\sum _{m=1}^n{\tan }^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$ is equal to

• A. ${\tan }^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$
• B. ${\tan }^{-1}\left(\frac{n^2+n}{n^2-n+2}\right)$
• C. ${\tan }^{-1}\left(\frac{n^2+n+2}{n^2+n}\right)$
• D. None of these

Answer 1

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

We have,
$\sum _{m=1}^n{\tan }^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$
$=\sum _{m=1}^n{\tan }^{-1}\left[\frac{2m}{1+(m^2+m+1)(m^2-m+1)}\right]$
$=\sum _{m=1}^n{\tan }^{-1}\left[\frac{(m^2+m+1)-(m^2-m+1)}{1+(m^2+m+1)(m^2-m+1)}\right]$
$=\sum _{m=1}^n\left\{{\tan }^{-1}\right[m^2+m+1]-{\tan }^{-1}[m^2-m+1\left]\right\}$
$={\tan }^{-1}3-{\tan }^{-1}1+{\tan }^{-1}7-{\tan }^{-1}3+({\tan }^{-1}13-{\tan }^{-1}7)+...+{\tan }^{-1}(n^2+n+1)-{\tan }^{-1}(n^2-n+1)$
$={\tan }^{-1}\left(\frac{n^2+n+1-1}{1+(n^2+n+1)\cdot 1}\right)={\tan }^{-1}\left(\frac{n^2+n}{2+n^2+n}\right)$