Prove that

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

Prove that

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestWhen \(n\) 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

Log in to reply

– Kartik Sharma · 2 months, 2 weeks ago

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

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

Log in to reply

– Mark Hennings · 2 months, 2 weeks ago

Very true!Log in to reply

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

Log in to reply

– Kartik Sharma · 2 months, 2 weeks ago

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

– A E · 2 months, 2 weeks ago

Ya I checked it .Log in to reply