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JEE MainMathematicsBinomial Theorem

If 1 0!40! + 1 2!38! + 1 4!36! + + 1 40!0! = 2^k m! , where k and m are positive integers, then the value of k+m is

Options

  1. A79
  2. B80
  3. C78
  4. D40

Correct answer

A. 79

Step-by-step solution

Let S = 1 0!40! + 1 2!38! + 1 4!36! + + 1 40!0! Multiplying and dividing the expression by 40! , we get: S = 1 40! [ 40! 0!40! + 40! 2!38! + 40! 4!36! + + 40! 40!0! ] S = 1 40! [ ⁴⁰C₀ + ⁴⁰C₂ + ⁴⁰C₄ + + ⁴⁰C₄₀ ] We know that the sum of even binomial coefficients is 2^ n-1 . For n=40 , this sum is 2⁴⁰⁻¹ = 2³⁹ . S = 1 40! 2³⁹ = 2³⁹ 40! Comparing this with 2^k m! , we get k = 39 and m = 40 . Therefore, k + m = 39 + 40 = 79 . Answer: 79

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