Q92 / 200 · Manipal (MET) 2014 · Mathematics
$\sum_{m=1}^n \tan ^{-1}\left(\frac{2 m}{m^4+m^2+2}\right)$ is equal to
- Correct answerA$\tan ^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$
- B$\tan ^{-1}\left(\frac{n^2-n}{n^2-n+2}\right)$
- C$\tan ^{-1}\left(\frac{n^2+n+2}{n^2+n}\right)$
- DNone of these
Solution
We have,
$\begin{aligned}
& \sum_{m=1}^n \tan ^{-1}\left(\frac{2 m}{m^4+m^2+2}\right) \\
&= \sum_{m=1}^n \tan ^{-1}\left[\frac{2 m}{1+\left(m^2+m+1\right)\left(m^2-m+1\right)}\right] \\
&= \sum_{m=1}^n \tan ^{-1}\left[\frac{\left(m^2+m+1\right)-\left(m^2-m+1\right)}{1+\left(m^2+m+1\right)\left(m^2-m+1\right)}\right] \\
&= \sum_{m=1}^n\left\{\tan ^{-1}\left[m^2+m+1\right]-\tan ^{-1}\left[m^2-m+1\right]\right\} \\
&= \tan ^{-1} 3-\tan ^{-1} 1+\tan ^{-1} 7-\tan ^{-1} 3 \\
&+\left(\tan ^{-1} 13-\tan ^{-1} 7\right)+\ldots+\tan ^{-1}\left(n^2+n+1\right) \\
&-\tan ^{-1}\left(n^2-n+1\right)
\end{aligned}$
$=\tan ^{-1}\left(\frac{n^2+n+1-1}{1+\left(n^2+n+1\right) \cdot 1}\right)=\tan ^{-1}\left(\frac{n^2+n}{2+n^2+n}\right)$
Hence, the correct answer is option: 'A'.