First slide

Properties of ITF

Question

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

Moderate

A

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

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

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

none of these

Solution

$\begin{array}{l} \tan^{- 1} \frac{2 m}{m^{4} + m^{2} + 2} \\ = \tan^{- 1} \frac{2 m}{1 + (m^{2} + m + 1) (m^{2} - m + 1)} \\ = \tan^{- 1} (m^{2} + m + 1) - \tan^{- 1} (m^{2} - m + 1) \\ so that \sum_{m = 1}^{n} \tan^{- 1} \frac{2 m}{m^{4} + m^{2} + 2} = (\tan^{1} 3 - \tan^{1} 1) + (\tan^{- 1} 7 - \tan^{- 1} 3) + \dots + [\tan^{- 1} (n^{2} + n + 1) - \tan^{- 1} (n^{2} - n + 1)] \\ = \tan^{- 1} (n^{2} + n + 1) - \tan^{- 1} 1 \\ = \tan^{- 1} \frac{n^{2} + n}{n^{2} + n + 2} \end{array}$

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