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If the sum of the infinite series _ n=1 ^ n^2 + kn + 1 (n+1)! is equal to 2e + 2 , then the value of the constant k is

Options

  1. A3
  2. B4
  3. C5
  4. D6

Correct answer

C. 5

Step-by-step solution

Let S = _ n=1 ^ n^2 + kn + 1 (n+1)! Let n+1 = m n = m-1 . The series starts from m=2 . The general term becomes: T_m = (m-1)^2 + k(m-1) + 1 m! = m^2 + (k-2)m + (2-k) m! Now, split the fraction: T_m = m (m-1)! + k-2 (m-1)! + 2-k m! Rewrite m as (m-1) + 1 in the first term: T_m = m-1+1 (m-1)! + k-2 (m-1)! + 2-k m! T_m = 1 (m-2)! + 1 (m-1)! + k-2 (m-1)! + 2-k m! T_m = 1 (m-2)! + k-1 (m-1)! + 2-k m! Summing from m=2 to : S = _ m=2 ^ 1 (m-2)! + (k-1) _ m=2 ^ 1 (m-1)! + (2-k) _ m=2 ^ 1 m! Using the standard exponential s

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