To prove a combinations equality:

(n+2 C r+1) = (n C r+1) + 2 (n C r ) + (n C r-1)

-> (n+2 C r+1) = (n+2)! / (n+2 - (r+1))! (r+1)!

-> (n C r+1) = n! / (n - (r+1))! (r+1)!

-> 2 (n C r ) = 2n! / (n - r)! r!

-> (n C r-1) = n! / (n - (r-1))! (r-1)!

Then I got this:

(n+2)! / (n - r + 1)! (r+1)!

= [n! / (n - r - 1)! (r+1)!] + [2n! / (n - r)! r!] + [n! / (n - r + 1)! (r-1)!]

Any help from here will be greatly appreciated. can get a common denominator, but I am wondering if there is a better/quicker approach.