JEE Main202127 Aug 2021Evening ShiftMathematicsSequences and SeriesActual
If 0 < x < 1 and y = 1 2 x 2 + 2 3 x 3 + 3 4 x 4 + … … , then the value of e 1 + y at x = 1 2 is:
Options
- A1 2 e 2
- B2 e
- C2 e 2
- D1 2 e
Correct answer
A. 1 2 e 2
Step-by-step solution
If 0 < x < 1 and y = 1 2 x 2 + 2 3 x 3 + 3 4 x 4 + … … y = 1 2 x 2 + 2 3 x 3 + 3 4 x 4 + … … y = 1 - 1 2 x 2 + 1 - 1 3 x 3 + 1 - 1 4 x 4 + ⋯ ⋯ · ⋯ = x 2 + x 3 + x 4 + ⋯ ⋯ - x 2 2 + x 3 3 + x 4 4 + ⋯ ⋯ ⋯ = x 2 1 - x + x - x + x 2 2 + x 3 3 + x 4 4 + ⋯ ⋯ ⋯ = x 1 - x + ℓ n ( 1 - x ) Put x = 1 2 y = 1 - ℓ n 2 Then, e 1 + y = e 1 + 1 - ℓ n 2 = 1 2 e 2