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Exercise 6.1 · Q6

Q.Show that (2n)!=2n⋅n![1.3.5...(2n−1)](2n)! = 2^n \cdot n! [1.3.5...(2n-1)]

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Prove the identity (2n)!=2n⋅n! [1⋅3⋅5⋯(2n−1)](2n)! = 2^n\cdot n!\,[1\cdot3\cdot5\cdots(2n-1)] by separating the 2n2n consecutive factors of (2n)!(2n)! into their odd-positioned and even-positioned parts.

By definition, (2n)!(2n)! is the product of the first 2n2n positive integers:

(2n)!=1×2×3×4×⋯×(2n−1)×(2n)(2n)! = 1\times2\times3\times4\times\cdots\times(2n-1)\times(2n)

Among these 2n2n consecutive integers, exactly nn are odd (1,3,5,…,2n−11,3,5,\ldots,2n-1) and exactly nn are even (2,4,6,…,2n2,4,6,\ldots,2n).

  1. Regroup the factors of (2n)!(2n)! by parity — since multiplication is commutative, we may reorder the product to list all odd factors first, then all even factors:

(2n)!=[1⋅3⋅5⋯(2n−1)]⏟n odd terms×[2⋅4⋅6⋯(2n)]⏟n even terms(2n)! = \underbrace{[1\cdot3\cdot5\cdots(2n-1)]}_{n\text{ odd terms}} \times \underbrace{[2\cdot4\cdot6\cdots(2n)]}_{n\text{ even terms}}

  1. Consider the even-term product 2⋅4⋅6⋯(2n)2\cdot4\cdot6\cdots(2n). Each even term can be written as 22 times an integer: 2=2(1)2=2(1), 4=2(2)4=2(2), 6=2(3)6=2(3), …, 2n=2(n)2n=2(n).
  2. Factor the common 22 out of each of the nn even terms:

2⋅4⋅6⋯(2n)=2(1)×2(2)×2(3)×⋯×2(n)=2n×(1×2×3×⋯×n)=2n n!2\cdot4\cdot6\cdots(2n) = 2(1)\times2(2)\times2(3)\times\cdots\times2(n) = 2^n\times(1\times2\times3\times\cdots\times n) = 2^n\,n!

  1. Substitute this back into the regrouped product from step 1: …

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