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JEE Previous Year Question

$If \sum_{r = 1}^{50} tan^{- 1} \frac{1}{2 r ^{2}} = p, then the value of tan p is :$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $\frac{101}{102}$
B $\frac{50}{51}$
C $100$
D $\frac{51}{50}$

Correct answer: B

Solution

(b) \sum tan^{- 1} (\frac{2}{4 r ^{2}}) = \sum tan^{- 1} (\frac{( 2 r + 1 ) - ( 2 r - 1 )}{1 + ( 2 r + 1 ) ( 2 r - 1 )}) \sum tan^{- 1} (2 r + 1) - tan^{- 1} (2 r - 1) tan^{- 1} (101) - tan^{- 1} 1 \Rightarrow tan^{- 1} \frac{50}{51}

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