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

Answer 1

Answer: (A)

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