How to find compact form of the sum $$\sum\limits_{k=0}^m (-1)^k \binom{n}{k} \binom{n}{m-k}$$
It looks like it's connected with Vandermonde's identity but I couldn't get to the solution.
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You mean the sum for the index $k$ and $i$ – marwalix Apr 04 '15 at 18:48
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You can write the sum in terms of hypergeometric function. – Marco Cantarini Apr 04 '15 at 18:59
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Note that, in $(1-x^2)^n=(1-x)^n(1+x)^n=\sum_{k=0}^{n}\binom{n}{k}(-x)^k\cdot \sum_{k=0}^{n}\binom{n}{k}x^k$. Consider the coefficient of $x^m$, for odd $m$, we get $\sum_{k=0}^n (-1)^k\binom{n}{k}\binom{n}{m-k}=0$, for even $m$, we get $\sum_{k=0}^n (-1)^k\binom{n}{k}\binom{n}{m-k}=(-1)^{\frac{m}{2}}\binom{n}{\frac{m}{2}}$.
Salomo
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Since $$ \sum_{k=0}^{n}\binom{n}{k}(-x)^k = (1-x)^n,\tag{1}$$ $$ \sum_{k=0}^{n}\binom{n}{k}x^k = (1+x)^n,\tag{2} $$ we have: $$ \sum_{k=0}^{n}\binom{n}{k}\binom{n}{m-k}(-1)^k = [x^m](1-x)^n(1+x)^n = [x^m](1-x^2)^n \tag{3}$$ hence:
$$ \sum_{k=0}^{n}\binom{n}{k}\binom{n}{m-k}(-1)^k = \left\{\begin{array}{rcl}0&\text{if}& m\text{ is odd}\\(-1)^{\frac{m}{2}}\binom{n}{\frac{m}{2}}&\text{if}& m\text{ is even.}\end{array}\tag{4}\right.$$
Jack D'Aurizio
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I don't quite understand the (3) step. How did you get this equation? Could you explain it a bit more? – VirrageS Apr 04 '15 at 19:29
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Very nice solution! I wonder if there exists a solution which uses only the binomial coefficients and/or their identities. – Hypergeometricx Apr 06 '15 at 15:15