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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