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Math Help - Proof

  1. #1
    t13
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    Lightbulb Proof

    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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  2. #2
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    Quote Originally Posted by t13 View Post
    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

    P(k+1)=\frac{(2k)!}{k!}\left(4k+2\right)=\frac{\le  ft(2[k+1]\right)!}{(k+1)!}

    \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)}

    =\frac{(2k+2)(2k+1)(2k)!}{(k+1)(2k)!}=\frac{(4k^2+  6k+2)(2k)!}{(k+1)k!}

    =\frac{(k+1)(4k+2)(2k)!}{(k+1)(2k)!}=\frac{(4k+2)(  2k)!}{k!}
    Last edited by Archie Meade; February 11th 2010 at 11:21 AM. Reason: typo
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  3. #3
    t13
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    perfect, thanks a ton!
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