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Math Help - Trick to evaluate this sum?

  1. #1
    Senior Member vincisonfire's Avatar
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    Trick to evaluate this sum?

    As my title says …
    Does someone know how to evaluate  \sum\limits_{k=0}^{2n}\frac{(-1)^k}{k!(2n-k)!}=0
    Thanks
    PS : I did not know where to post this but it arose from an algebra problem so …
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  2. #2
    MHF Contributor

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    \sum_{k=0}^{2n} \frac{(-1)^k}{k!(2n-k)!}=\frac{1}{(2n)!} \sum_{k=0}^{2n} \frac{(-1)^k(2n)!}{k!(2n-k)!}=\frac{1}{(2n)!} \sum_{k=0}^{2n} (-1)^k \binom{2n}{k} = \frac{1}{(2n)!}(1-1)^{2n}=0.
    Last edited by mr fantastic; September 16th 2010 at 08:49 PM.
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  3. #3
    Senior Member vincisonfire's Avatar
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    Indeed, I might as well ask what is 1+1. I should definitely get back to it.
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