You can
put this solution on YOUR website!The notation
tells you to simply multiply that number by every integer that is less than that number until you hit 1. So if
, this means that
.
In this case,
!=(2n+2)(2n+2-1)(2n+2-2)(2n+2-3)(2n+2-4)\ldots(3)(2)(1))
which simplifies to
!=(2n+2)(2n+1)(2n)(2n-1)(2n-2)\ldots(3)(2)(1))
and
!=(2n)(2n-1)(2n-2)(2n-3)\ldots(3)(2)(1))
So
!}{(2n)!}=\frac{(2n+2)(2n+1)(2n)(2n-1)(2n-2)\ldots(3)(2)(1)}{(2n)(2n-1)(2n-2)(2n-3)\ldots(3)(2)(1)})
Highlight the common terms.
Cancel out the common terms.
!}{(2n)!}=(2n+2)(2n+1))
Simplify.
!}{(2n)!}=4n^2+6n+2)
FOIL
to get
(