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)!.
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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)!. What you need to prove is whether or not
Last edited by Archie Meade; February 11th 2010 at 12:21 PM. Reason: typo
perfect, thanks a ton!
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