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AP EAMCET201921 Apr 2019Evening ShiftMathematicsLimitsActual

lim n → ∞ 1 3 . 7 + 1 7 . 11 + 1 11 . 15 + … + ( n terms ) =

Options

  1. A1 12
  2. B1 4
  3. C1 3
  4. D0

Correct answer

A. 1 12

Step-by-step solution

The series is given as, = lim n → ∞ 1 3 · 7 + 1 7 · 11 + 1 11 · 15 + … + n   terms = lim n → ∞ 1 4 4 3 · 7 + 4 7 · 11 + 4 11 · 15 … = lim n → ∞ 1 4 1 3 - 1 7 + 1 7 - 1 11 + 1 11 - 1 15 + … 1 n + 1 n - 1 n + 4 = lim n → ∞ 1 4 1 3 - 1 n + 4 = lim n → ∞ 1 12 - lim n → ∞ 1 4 ( n + 4 ) = 1 12 - 0 = 1 12

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