summation evaluation

**pandakrap** · November 4th 2008, 09:11 AM

could I get some help with evaluating this sum.

**PaulRS** · November 4th 2008, 09:36 AM

Note that your sum is equal to: $ \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 $x^{2008}$ in the product: $ \left( {1 - x} \right)^{2009} \cdot \left( {1 + x} \right)^{2008} $ ( use the Binomial Theorem)

On the other hand: $ \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: $ \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 $x^{2008}$ as $\binom{2008}{1004}$

**pandakrap** · November 4th 2008, 10:18 AM

thanks a bunch

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