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JEE MainMathematicsBinomial Theorem

The value of _ k=1 ²⁰ 2k (2k)!(40-2k)! is

Options

  1. A2³⁹ 40!
  2. B2³⁸ 39!
  3. C2³⁸ 40!
  4. D2³⁹ 39!

Correct answer

B. 2³⁸ 39!

Step-by-step solution

Let S = _ k=1 ²⁰ 2k (2k)!(40-2k)! We can simplify the term in the denominator as (2k)! = 2k (2k-1)! . Substituting this into the series, we get: S = _ k=1 ²⁰ 2k 2k (2k-1)!(40-2k)! = _ k=1 ²⁰ 1 (2k-1)!(40-2k)! Notice that the sum of the numbers in the factorials in the denominator is (2k-1) + (40-2k) = 39 . Multiplying and dividing the expression by 39! , we get: S = 1 39! _ k=1 ²⁰ 39! (2k-1)!(39 - (2k-1))! S = 1 39! _ k=1 ²⁰ ³⁹C_ 2k-1 Expanding the summation by substituting k = 1, 2, , 20 , we get: S = 1 39! [ ³⁹C₁

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