If S is the sum of the first 10 terms of the series $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots .$, then $\tan (\mathrm{S})$ is equal to:

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Inverse Trigonometric Functions

If S is the sum of the first 10 terms of the series

tan^{- 1} (\frac{1}{3}) + tan^{- 1} (\frac{1}{7}) + tan^{- 1} (\frac{1}{13}) + tan^{- 1} (\frac{1}{21}) + \dots .

, then

tan (S)

is equal to:

Answer: (A)

The denominators

3, 7, 13, 21, \dots

are

n^{2} + n + 1

, and

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

, so each term equals

tan^{- 1} (n + 1) - tan^{- 1} n

. The sum telescopes, and the first ten terms give

S = tan^{- 1} 11 - tan^{- 1} 1

. Therefore

tan S = \frac{11 - 1}{1 + 11 \cdot 1} = \frac{10}{12} = \frac{5}{6}

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option

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