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Let f: N N be a function satisfying f(x+y) = f(x)f(y) for all x, y N . If f(1) = 4 , then the value of _ k=1 ^ 2k f(k) is equal to

Options

  1. A8 9
  2. B4 9
  3. C2 3
  4. D32 9

Correct answer

A. 8 9

Step-by-step solution

Given f(x+y) = f(x)f(y) for all x, y N and f(1) = 4 . By substituting x=1, y=1 , we get f(2) = (f(1))^2 = 4^2 . By induction, f(k) = 4^k for all k N . The given series is S = _ k=1 ^ 2k f(k) = 2 _ k=1 ^ k ( 1 4 )^k . Let S₁ = _ k=1 ^ k ( 1 4 )^k = 1 4 + 2 ( 1 4 )^2 + 3 ( 1 4 )^3 + Multiply by 1 4 : 1 4 S₁ = ( 1 4 )^2 + 2 ( 1 4 )^3 + Subtracting the two equations: S₁ - 1 4 S₁ = 1 4 + ( 1 4 )^2 + ( 1 4 )^3 + 3 4 S₁ = 1 4 1 - 1 4 = 1 3 S₁ = 4 9 Therefore, the required sum is S = 2S₁ = 2 4 9 = 8 9 . Answer: 8 9

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