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Math Help - interesting combinatorial identity proof

  1. #1
    Eater of Worlds
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    interesting combinatorial identity proof

    Hi ALL;

    Here is one I found interesting. If you want to give it a go.

    I am sorry, I can not get those capital Pi's (product series) to display larger.

    Prove for integers n and t with n\geq 0, \;\ t\geq 1

    \frac{1}{t!}\sum_{k=0}^{n}\prod_{j=1}^{t}(n-k+j)=\frac{1}{(t+1)!}\prod_{j=1}^{t+1}(n+j)

    In order to prove this, show the following. The first two can be done algebraically and the last inductively on n.

    \frac{1}{t!}\sum_{k=0}^{n}\prod_{j=1}^{t}(n-k+j)=\sum_{k=0}^{n}\binom{t+k}{k}

    \frac{1}{(t+1)!}\prod_{j=1}^{t+1}(n+j)=\binom{t+n+  1}{n}

    \sum_{k=0}^{n}\binom{t+k}{k}=\binom{t+n+1}{n}
    Last edited by Chris L T521; November 29th 2009 at 04:35 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by galactus View Post
    Hi ALL;

    Here is one I found interesting. If you want to give it a go.

    I am sorry, I can not get those capital Pi's (product series) to display larger.

    Prove for integers n and t with n\geq 0, \;\ t\geq 1

    \frac{1}{t!}\sum_{k=0}^{n}{\Pi}_{j=1}^{t}(n-k+j)=\frac{1}{(t+1)!}{\Pi}_{j=1}^{t+1}(n+j)

    In order to prove this, show the following. The first two can be done algebraically and the last inductively on n.

    \frac{1}{t!}\sum_{k=0}^{n}\Large{\Pi}_{j=1}^{t}(n-k+j)=\sum_{k=0}^{n}\binom{t+k}{k}

    \frac{1}{(t+1)!}{\Pi}_{j=1}^{t+1}(n+j)=\binom{t+n+  1}{n}

    \sum_{k=0}^{n}\binom{t+k}{k}=\binom{t+n+1}{n}
    Hey galactus,

    You can use \prod to generate \prod. I've edited your post and put them in there for you.
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