How do you simplify #((2n+3)!)/((2n)!)#?

1 Answer
Alan P.
Nov 14, 2015

#((2n+3)!)/((2n)!) = (2n+3)(2n+2)(2n+1)#

Explanation:

#((2n+3)!)/((2n)!)#

#color(white)("XX") = ((2n+3)xx(2n+2)xx(2n+1)xxcancel((2n))xxcancel((2n-1))xxcancel((2n-2))xx...xxcancel((1)))/(cancel((2n))xxcancel((2n-1))xxcancel((2n-2))xx...xxcancel((1)))#

#color(white)("XX")=((2n+3)(2n+2)(2n+1)#

Related questions
See all questions in Factorial Identities
Impact of this question
13414 views around the world
You can reuse this answer
Creative Commons License