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If S_n = 1 2 3! + 2 2^2 4! + 3 2^3 5! + upto n terms then the sum of the infinite terms is:

Options

  1. A1
  2. B2 3
  3. Ce
  4. D4

Correct answer

B. 2 3

Step-by-step solution

The r^ th term is T_r = r 2^r (r+2)! . We can rewrite the numerator: r 2^r = (r+2-2)2^r = (r+2)2^r - 2^ r+1 . So, T_r = (r+2)2^r (r+2)! - 2^ r+1 (r+2)! = 2^r (r+1)! - 2^ r+1 (r+2)! . Now, calculate the sum S_n = _ r=1 ^n T_r : T₁ = 2^1 2! - 2^2 3! T₂ = 2^2 3! - 2^3 4! T_n = 2^n (n+1)! - 2^ n+1 (n+2)! Summing these terms (telescoping series): S_n = 2^1 2! - 2^ n+1 (n+2)! = 1 - 2^ n+1 (n+2)! . As n , the term 2^ n+1 (n+2)! 0 . Thus, S_ = 1 - 0 = 1 .

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