Question

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

Consider, ${\tan }^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$
$={\tan }^{-1}\left(\frac{2m}{1+(m^4+2m^2-m^2+1)}\right)$
$={\tan }^{-1}\left(\frac{2m}{1+(m^2+1{)}^2-m^2}\right)$
$={\tan }^{-1}\left(\frac{2m}{1+(m^2+1+m)(m^2+1-m)}\right)$
$={\tan }^{-1}(m^2+1+m)-{\tan }^{-1}(m^2-m+1)$
Now, ${\sum }_{m=1}^n{\tan }^{-1}\left(\frac{2m}{m^4+m^2+2}\right)={\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(7\right)-{\tan }^{-1}\left(3\right)+......{\tan }^{-1}(n^2+1+n)-{\tan }^{-1}(n^2-n+1)$
$={\tan }^{-1}(n^2+n+1)-{\tan }^{-1}\left(1\right)$
$={\tan }^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$