Question

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

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

Answer 1

Answer: (A)

We have $\sum _{m=1}^n{\tan }^{-1}\frac{2m}{m^4+m^2+2}=\sum _{m=1}^n$
${\tan }^{-1}\frac{2m}{1+m^2+m+1m^2-m+1}$
$=\sum _{m=1}^n{\tan }^{-1}\frac{m^2+m+1-m^2-m+1}{1+m^2+m+1-m^2-m+1}$
$=\sum _{m=1}^n{\tan }^{-1}{m}^2+m+1-{\tan }^{-1}{m}^2-m+1$
$={\tan }^{-1}3-{\tan }^{-1}1+{\tan }^{-1}7-{\tan }^{-1}3+{\tan }^{-1}13-{\tan }^{-1}7+\dots +{\tan }^{-1}{n}^2+n+1-{\tan }^{-1}{n}^2-n+1$
$={\tan }^{-1}{n}^2+n+1-{\tan }^{-1}1={\tan }^{-1}$
$\frac{n^2+n}{2+n^2+n}$