∑ _ m = 1 ^ n ^ - 1 ( 2m m^ 4 + m^ 2 + 2 )is equal to

We have, $\mathop{∑}_{m = 1}^{n}\tan^{- 1}\left(\dfrac{2m}{{m}^{4} + {m}^{2} + 2}\right)$

$= \mathop{∑}_{m = 1}^{n}\tan^{- 1}\left(\dfrac{\left({m}^{2} + m + 1\right) - \left({m}^{2} - m + 1\right)}{1 + \left({m}^{2} + m + 1\right)\left({m}^{2} - m + 1\right)}\right)$

$= \mathop{∑}_{m = 1}^{n}\left[\tan^{- 1}\left({m}^{2} + m + 1\right) - \tan^{- 1}\left({m}^{2} - m + 1\right)\right]$

= \left(\tan^{- 1}3 - \tan^{- 1}1\right) + \left(\tan^{- 1}7 - \tan^{- 1}3\right) + \left(\tan^{- 1} ( 13 ) - \tan^{- 1}7\right) + \ldots + \left(\tan^{- 1}\left({n}^{2} + n + 1\right)\right\left- \tan^{- 1}\left({n}^{2} - n + 1\right)\right)

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

$= \tan^{- 1}\left(\dfrac{{n}^{2} + n + 1 - 1}{1 + \left({n}^{2} + n + 1\right)}\right) = \tan^{- 1}\left(\dfrac{{n}^{2} + n}{{n}^{2} + n + 2}\right)$