summation evaluation

• November 4th 2008, 09:11 AM
pandakrap
summation evaluation
could I get some help with evaluating this sum.
• November 4th 2008, 09:36 AM
PaulRS
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 $\binom{2008}{1004}$
• November 4th 2008, 10:18 AM
pandakrap
:)
thanks a bunch