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Thread: summation evaluation

  1. #1
    Junior Member
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    summation evaluation

    could I get some help with evaluating this sum.
    Attached Thumbnails Attached Thumbnails summation evaluation-sum.bmp  
    Last edited by pandakrap; Nov 4th 2008 at 10:18 AM.
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  2. #2
    Super Member PaulRS's Avatar
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    Note that your sum is equal to: $\displaystyle
    \sum\limits_{k = 0}^{2008}{\left( { - 1} \right)^k \cdot \binom{2009}{k}\cdot{\binom{2008}{2008-k}}}
    $

    And that corresponds to the coefficient of $\displaystyle x^{2008}$ in the product: $\displaystyle
    \left( {1 - x} \right)^{2009} \cdot \left( {1 + x} \right)^{2008}
    $ ( use the Binomial Theorem)

    On the other hand: $\displaystyle
    \left( {1 - x} \right)^{2009} \cdot \left( {1 + x} \right)^{2008} = \left( {1 - x} \right) \cdot \left( {1 - x^2 } \right)^{2008}
    $

    By the Binomial Theorem:$\displaystyle
    \left( {1 - x} \right) \cdot \left( {1 - x^2 } \right)^{2008} = \left( {1 - x} \right) \cdot \sum\limits_{k = 0}^{2008} {\binom{2008}{k} \cdot \left( { - x^2 } \right)^k }
    $

    So we recognise the coefficient of $\displaystyle x^{2008}$ as $\displaystyle \binom{2008}{1004}$
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  3. #3
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    :)

    thanks a bunch
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