Question

$\sum _{m-1}^n{\tan }^{-1}\left(\frac{2m}{m^v+m^2+2}\right)$ 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. $-{\cot }^{-1}\left(\frac{n^2+n+2}{n^2+n}\right)$

Answer 1

The correct option is A

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

$\sum _{m_1}^n{\tan }^{-1}\frac{2m}{m^4+m^2+2}=\sum _{m-1}^n{\tan }^{-1}\left(\frac{2m}{1+(m^2+m+1)(m^2-m+1)}\right)$

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

$=\sum _{m-1}n\left[ta{n}^{-1}\right(m^2+m+1)-tan-1(m^2-m+1\left)\right]$

$=({\tan }^{-1}3-{\tan }^{-1}1)+({\tan }^{-1}7-{\tan }^{-1}3)+({\tan }^{-1}13-{\tan }^{-1}7)+⋮⋮+\left[{\tan }^{-1}\right(n^2+n+1)-{\tan }^{-1}(n^2-n+1)$

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