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)!.
What you need to prove is whether or not
$\displaystyle P(k+1)=\frac{(2k)!}{k!}\left(4k+2\right)=\frac{\le ft(2[k+1]\right)!}{(k+1)!}$
$\displaystyle \frac{\left(2[k+1]\right)!}{(k+1)!}=\frac{(2k+2)!}{(k+1)!}=\frac{(2k +2)(2k+1)(2k)(2k-1)(2k-2)(2k-3)....3(2)}{(k+1)(k)(k-1)(k-2)(k-3)....3(2)}$
$\displaystyle =\frac{(2k+2)(2k+1)(2k)!}{(k+1)(2k)!}=\frac{(4k^2+ 6k+2)(2k)!}{(k+1)k!}$
$\displaystyle =\frac{(k+1)(4k+2)(2k)!}{(k+1)(2k)!}=\frac{(4k+2)( 2k)!}{k!}$