Show that (2n choose n+1) + (2n choose n) = (2n+2 choose n+1) / 2
2n choose n+1 = 2n!/(((n+1)!)((2n-(n+1)!)) = 2n! / (((n+1)!)((n-1)!))
2n choose n = 2n!/(n! (2n-n)!) = 2n! / (n! * n!)
is there a way to simplify these two expressions before getting a common denominator and adding them.
This is what I got with out simplifying and then adding them:
[(2n! * n! * n!) + (2n! * (n+1)! (n-1)!)] / [ (n+1)! * (n-1)! * n! * n!]
And I got stuck again trying to simplify that.