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\( \sum_{m=1}^{n} \tan^{-1}\left(\frac{2m}{m^4 + m^2 + 2}\right) \) is equal to
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If expression looks complex, try converting into telescoping form.
MET - 2014
MET
Updated On:
Apr 23, 2026
$\tan^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$
$\tan^{-1}\left(\frac{n^2-n}{n^2-n+2}\right)$
$\tan^{-1}\left(\frac{n^2+n+2}{n^2+n}\right)$
None of these
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