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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
5 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 - 5 months, 2 weeks ago

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Yeah. It is Dixon's formula for \(F(a,b,c;1+c-a,1+c-b;1)\).

Kartik Sharma - 5 months, 2 weeks ago

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The sum should start from \(0\).

Ishan Singh - 5 months, 2 weeks ago

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Very true!

Mark Hennings - 5 months, 2 weeks ago

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Seems interesting and damn tough!!!!

A E - 5 months, 3 weeks ago

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Mark Hennings has provided a method to do it. You can check.

Kartik Sharma - 5 months, 2 weeks ago

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Ya I checked it .

A E - 5 months, 2 weeks ago

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