How do you simplify 2^n/[(2n)!]?

1 Answer
Alan N.
Apr 29, 2016

1/((2n-1)!! * n!)

Explanation:

2^n/((2n)!) = (2*2*2*2* ....) / [2n* (2n-1) *(2n-2) ....4*3* 2*1]

Notice that each of the n 2's in the numerator will cancel with each of the n even terms in the denominator.

2^n/((2n)!) = cancel(2*2*2*2* ....) / [cancel(2)n *(2n-1) *cancel(2) (n-1) ....cancel(2)2*3* cancel(2)1*1]

= 1/(n*(2n-1)*(n-1) ...2*3*1*1)

Rearranging terms
= 1/{[(2n-1)*(2n-3) ....3*1] * [n*(n-1)*(n-2) .... 2*1]

The 1st term in the denominator is the product of odd integers from (2n-1) to 1. This is the known as the double factorial (2n-1)!!. The 2nd term is simply n!

Hence:

2^n/((2n)!) = 1/((2n-1)!! * n!)

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