∑m=1n tan−1⁡2mm4+m2+2=

∑m=1n tan−1⁡2mm4+m2+2 =∑m=1n tan−1⁡2m1+m4+m2+1=∑m=1n tan−1⁡2m1+m2+m+1m2−m+1 ∑m=1n tan−1⁡m2+m+1−m2−m+11+m2+m+1m2−m+1 =∑m=1n tan−1m2+m+1- tan−1m2−m+1 = tan−1n2+n+1-tan-11=tan-1n2+n+1-11+n2+n+11=tan−1⁡n2+nn2+n+2

∑m=1n tan−1⁡2mm4+m2+2=

∑m=1n tan−1⁡2mm4+m2+2 =∑m=1n tan−1⁡2m1+m4+m2+1=∑m=1n tan−1⁡2m1+m2+m+1m2−m+1 ∑m=1n tan−1⁡m2+m+1−m2−m+11+m2+m+1m2−m+1 =∑m=1n tan−1m2+m+1- tan−1m2−m+1 = tan−1n2+n+1-tan-11=tan-1n2+n+1-11+n2+n+11=tan−1⁡n2+nn2+n+2

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$\sum_{m = 1}^{n} \tan^{- 1} \frac{2 m}{m^{4} + m^{2} + 2} =$

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$\tan^{- 1} (n^{2} - n + 1)$

$\tan^{- 1} (\frac{n^{2} + n}{n^{2} + n + 2})$

$\tan^{- 1} (n^{2} + n)$

$\tan^{- 1} (n^{2} + n + 1)$

answer is C.

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Detailed Solution

$\sum_{m = 1}^{n} \tan^{- 1} \frac{2 m}{m^{4} + m^{2} + 2} = \sum_{m = 1}^{n} \tan^{- 1} \frac{2 m}{1 + (m^{4} + m^{2} + 1)} = \sum_{m = 1}^{n} \tan^{- 1} \frac{2 m}{1 + (m^{2} + m + 1) (m^{2} - m + 1)} \sum_{m = 1}^{n} \tan^{- 1} (\frac{(m^{2} + m + 1) - (m^{2} - m + 1)}{1 + (m^{2} + m + 1) (m^{2} - m + 1)}) = \sum_{m = 1}^{n} \tan^{- 1} (m^{2} + m + 1) - \tan^{- 1} (m^{2} - m + 1) = \tan^{- 1} (n^{2} + n + 1) - \tan^{- 1} 1 = \tan^{- 1} (\frac{n^{2} + n + 1 - 1}{1 + (n^{2} + n + 1) 1}) = \tan^{- 1} (\frac{n^{2} + n}{n^{2} + n + 2})$