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The value of $\sum\limits_{m=1}^{n}\tan^{-1}(\frac{2m}{m^4+m^2+2})$ is: |
$\tan^{-1}(n^2+n)$ $\tan^{-1}(\frac{n^2+n+1}{n^2+n+2})$ $\tan^{-1}(\frac{n}{n+1})$ $\tan^{-1}(\frac{n^2+n}{n^2+n+2})$ |
$\tan^{-1}(\frac{n^2+n}{n^2+n+2})$ |
$\sum\limits_{m=1}^{n}\tan^{-1}[\frac{2m}{1+(m^2+m+1).(m^2-m+1)}]$ $\sum\limits_{m=1}^{n}(\tan^{-1}(m^2+m+1)-\tan^{-1}(m^2-m+1))=\tan^{-1}(n^2+n+1)-\tan^{-1}(1)=\tan^{-1}(\frac{n^2+n}{2+n+n^2})$ |
