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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. Mark Hennings · 2 months, 2 weeks ago

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@Mark Hennings Yeah. It is Dixon's formula for \(F(a,b,c;1+c-a,1+c-b;1)\). Kartik Sharma · 2 months, 2 weeks ago

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The sum should start from \(0\). Ishan Singh · 2 months, 2 weeks ago

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@Ishan Singh Very true! Mark Hennings · 2 months, 2 weeks ago

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

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

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@Kartik Sharma Ya I checked it . A E · 2 months, 2 weeks ago

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