QJEEM21Maa31e: JEE Main 2021 Mathematics - Class 12 JEEM21Maa31e Solution: If ∑ r = 1 50 tan − 1 1 2 r 2 = p , then the value of tan p is :

Question

Verified Solution • 2025–26 Session moderate

If

\sum_{r = 1}^{50} \tan^{- 1} \frac{1}{2 r^{2}} = p

, then the value of tan p is :

Accepted Answer

\sum_{r = 1}^{50} \tan^{- 1} (\frac{2}{4 r^{2}}) = \sum_{r = 1}^{50} \tan^{- 1} (\frac{(2 r + 1) - (2 r - 1)}{1 + (2 r + 1) (2 r - 1)})

=

\sum_{r = 1}^{50} \tan^{- 1} (2 r + 1) - \tan^{- 1} (2 r - 1)

=

\tan^{- 1} (101) - \tan^{- 1} 1 = \tan^{- 1} \frac{50}{51}

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