How do you simplify #((2n+2)!)/((2n)!)#?

1 Answer
Gió
Nov 24, 2015

I found: #(2n+2)(2n+1)#

Explanation:

Remembering that we can write:
#n! =n*(n-1)!#

we have (applying twice the above property to the numerator):
#((2n+2)!)/((2n)!) =((2n+2)(2n+1)cancel((2n)!))/(cancel((2n)!))=(2n+2)(2n+1)#

Check with #n=3#
#((2n+2)!)/((2n)!) =(8!)/(6!)=40320/720=56#
and:
#(2n+2)(2n+1)=8*7=56#

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