The value of $\sum\limits_{m=1}^{n}\tan^{-1}(\frac{2m}{m^4+m^2+2})$ is:
Answer & explanation
Correct answer: option 4
$\sum\limits_{m=1}^{n}\tan^{-1}[\frac{2m}{1+(m^2+m+1).(m^2-m+1)}]$
$\sum\limits_{m=1}^{n}(\tan^{-1}(m^2+m+1)-\tan^{-1}(m^2-m+1))=\tan^{-1}(n^2+n+1)-\tan^{-1}(1)=\tan^{-1}(\frac{n^2+n}{2+n+n^2})$