JEE Main202116 Mar 2021Morning ShiftMathematicsInverse Trigonometric FunctionsActual
Let S k = ∑ r = 1 k tan - 1 6 r 2 2 r + 1 + 3 2 r + 1 , then lim k → ∞ S k is equal to :
Options
- Atan - 1 3 2
- Bπ 2
- Ccot - 1 3 2
- Dtan - 1 3
Correct answer
C. cot - 1 3 2
Step-by-step solution
Given S k = ∑ r = 1 k tan - 1 6 r 2 2 r + 1 + 3 2 r + 1 ⇒ S k = ∑ r = 1 k tan - 1 2 r · 3 r 2 2 r · 2 + 3 2 r · 3 Divide by 3 2 r we get, S k = ∑ r = 1 k tan - 1 2 3 r 2 3 2 r . 2 + 3 S k = ∑ r = 1 k tan - 1 2 3 r 3 2 3 2 r + 1 + 1 Let 2 3 r = t S k = ∑ r = 1 k tan - 1 t 3 1 + 2 3 t 2 S k = ∑ r = 1 k tan - 1 t - 2 t 3 1 + t . 2 t 3 S k = ∑ r = 1 k tan - 1 t - tan - 1 2 t 3 S k = ∑ r = 1 k tan - 1 2 3 r - tan - 1 2 3 r + 1 S k = tan - 1 2 3 - tan -