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  4. displaystyle ∑ m = 1n( tan)- 1((2 m/m4 + m2 + 2)) is equal to

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Q. $\displaystyle \sum _{m = 1}^{n}\left(\tan\right)^{- 1}\left(\frac{2 m}{m^{4} + m^{2} + 2}\right)$ is equal to

NTA AbhyasNTA Abhyas 2022

Solution:

We have $\displaystyle\sum_{m=1}^n \tan ^{-1} \frac{2 m}{m^4+m^2+2}=\sum_{m=1}^n$
$\tan ^{-1} \frac{2 m}{1+m^2+m+1 m^2-m+1} $
$=\displaystyle\sum_{m=1}^n \tan ^{-1} \frac{m^2+m+1-m^2-m+1}{1+m^2+m+1-m^2-m+1} $
$=\displaystyle\sum_{m=1}^n\tan ^{-1} m^2+m+1-\tan ^{-1} m^2-m+1 $
$=\tan ^{-1} 3-\tan ^{-1} 1+\tan ^{-1} 7-\tan ^{-1} 3+ \tan ^{-1} 13-\tan ^{-1} 7+\ldots+ \tan ^{-1} n^2+n+1-\tan ^{-1} n^2-n+1$
$=\tan ^{-1} n^2+n+1-\tan ^{-1} 1=\tan ^{-1} $
$\frac{n^2+n}{2+n^2+n}$