Originally Posted by
t13 Having trouble with a proof here.
If n is a natural number, then
2 * 6 * 10 * 14 * ... * (4n-2) = (2n)!/(n!).
I came up with the following so far:
P(k+1) = [(2n)!/(n!)] * (4k+2).
Not really sure how to make this equal [2(k+1)]!/(k+1)!.