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JEE Main202228 Jun 2022Evening ShiftMathematicsInverse Trigonometric FunctionsActual

The value of lim n → ∞ 6 tan ∑ r = 1 n tan - 1 1 r 2 + 3 r + 3 is equal to

Options

  1. A1
  2. B2
  3. C3
  4. D6

Correct answer

C. 3

Step-by-step solution

Let S = ∑ r = 1 n tan - 1 1 r 2 + 3 r + 3 So, T r = tan - 1 1 r 2 + 3 r + 2 + 1 = tan - 1 ( r + 2 ) - ( r + 1 ) 1 + ( r + 1 ) ( r + 2 ) = tan - 1 r + 2 - tan - 1 r + 1 T 1 = tan - 1 3 - tan - 1 2 T 2 = tan - 1 4 - tan - 1 3 . . T n = tan - 1 n + 2 - tan - 1 n + 1 ⇒ ∑ r = 1 n tan - 1 1 r 2 + 3 r + 3 = tan - 1 n + 2 - tan - 1 2 i.e. 6 tan lim n → ∞ ∑ r = 1 n tan - 1 1 r 2 + 3 r + 3 = 6 tan lim n → ∞ ∑ r = 1 n tan - 1 n + 2 - tan - 1 2 = 6 tan π 2 - tan - 1 2

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