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Math Help - 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; November 4th 2008 at 11:18 AM.
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  2. #2
    Super Member PaulRS's Avatar
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    Note that your sum is equal to: <br />
\sum\limits_{k = 0}^{2008}{\left( { - 1} \right)^k  \cdot \binom{2009}{k}\cdot{\binom{2008}{2008-k}}} <br />

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

    On the other hand: <br />
\left( {1 - x} \right)^{2009}  \cdot \left( {1 + x} \right)^{2008}  = \left( {1 - x} \right) \cdot \left( {1 - x^2 } \right)^{2008} <br />

    By the Binomial Theorem: <br />
\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 } <br />

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

    thanks a bunch
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