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

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Inverse Trigonometric Functions

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

, then the value of

tan p

is:

Answer: (D)

Write the general term so that it telescopes:

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

Since

2 r + 1 > 2 r - 1 \geq 1 > 0

for

r \geq 1

, this gives

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

. Summing from

r = 1

50

, all interior terms cancel and

p = tan^{- 1} (101) - tan^{- 1} (1)

Therefore

tan p = \frac{101 - 1}{1 + 101 \cdot 1} = \frac{100}{102} = \frac{50}{51}

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