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

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a

tan−1⁡n2+n+1

b

tan−1⁡n2−n+1

c

tan−1⁡n2+nn2+n+2

d

none of these

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detailed solution

Correct option is C

tan−1⁡2mm4+m2+2=tan−1⁡2m1+m2+m+1m2−m+1=tan−1⁡m2+m+1−tan−1⁡m2−m+1 so that ∑m=1n tan−1⁡2mm4+m2+2=tan1⁡3−tan1⁡1+tan−1⁡7−tan−1⁡3+…+tan−1⁡n2+n+1−tan−1⁡n2−n+1=tan−1⁡n2+n+1−tan−1⁡1=tan−1⁡n2+nn2+n+2

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