Question 149898
The notation {{{n!}}} tells you to simply multiply that number by every integer that is less than that number until you hit 1. So if {{{n=5}}}, this means that {{{5!=5(5-1)*(5-2)*(5-3)*(5-4)=5(4)(3)(2)(1)=120}}}.

In this case, *[Tex \LARGE (2n+2)!=(2n+2)(2n+2-1)(2n+2-2)(2n+2-3)(2n+2-4)\ldots(3)(2)(1)]

which simplifies to *[Tex \LARGE (2n+2)!=(2n+2)(2n+1)(2n)(2n-1)(2n-2)\ldots(3)(2)(1)]

and *[Tex \LARGE (2n)!=(2n)(2n-1)(2n-2)(2n-3)\ldots(3)(2)(1)]

So *[Tex \LARGE \frac{(2n+2)!}{(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)}]

<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/highlighted7-27-08.png" alt="Photobucket - Video and Image Hosting"> Highlight the common terms.

<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/canceled7-27-08.png" alt="Photobucket - Video and Image Hosting"> Cancel out the common terms.

*[Tex \LARGE \frac{(2n+2)!}{(2n)!}=(2n+2)(2n+1)] Simplify.

*[Tex \LARGE \frac{(2n+2)!}{(2n)!}=4n^2+6n+2] FOIL {{{(2n+2)(2n+1)}}} to get {{{4n^2+4n+2n+2=4n^2+6n+2}}}