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# Binomial Coefficient Challenge 1!

Prove that

$\sum_{k=0}^n {\binom{n}{k}^3 {(-1)}^k} = \cos\left(\frac{\pi n}{2}\right) \binom{n}{n/2}\binom{3n/2}{n}$

Note by Kartik Sharma
2 months, 3 weeks ago

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When $$n$$ is odd, replacing $$k$$ with $$n-k$$ shows that the sum is zero. The result for even $$n$$ is the original Dixon's identity. · 2 months, 2 weeks ago

Yeah. It is Dixon's formula for $$F(a,b,c;1+c-a,1+c-b;1)$$. · 2 months, 2 weeks ago

The sum should start from $$0$$. · 2 months, 2 weeks ago

Very true! · 2 months, 2 weeks ago

Seems interesting and damn tough!!!! · 2 months, 3 weeks ago

@A E Mark Hennings has provided a method to do it. You can check. · 2 months, 2 weeks ago