Explanation

(a)
We have, $\sum_{m = 1}^{n}{\tan^{- 1}\left( \frac{2m}{m^{4} + m^{2} + 2} \right)}$
$= \sum_{m = 1}^{n}{\tan^{- 1}\left( \frac{2m}{1 + \left( m^{2} + m + 1 \right)(m^{2} - m + 1)} \right)}$
$= \sum_{m = 1}^{n}{\tan^{- 1}\left( \frac{\left( m^{2} + m + 1 \right) - (m^{2} - m + 1)}{1 + \left( m^{2} + m + 1 \right)(m^{2} - m + 1)} \right)}$
$= \sum_{m = 1}^{n}{\lbrack\tan^{- 1}{\left( m^{2} + m + 1 \right) - \tan^{- 1}{(m^{2} - m + 1)\rbrack}}}$
$= (\tan^{- 1}{3 - {\tan^{- 1}1)}{+ ({\tan^{- 1}7}{- {\tan^{- 1}3)} +}}}$
$({\tan^{- 1}13}{- \tan^{- 1}{7) + ... + \ }}$
$\lbrack\tan^{- 1}{\left( n^{2} + n + 1 \right) - \tan^{- 1}{\left( n^{2} - n + 1 \right)\rbrack}}$
$= \tan^{- 1}{\frac{n^{2} + n + 1 - 1}{1 + (n^{2} + n + 1) \bullet 1} = \tan^{- 1}\left( \frac{n^{2} + n}{2 + n^{2} + n} \right)}$