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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 9 months ago

Easy Math Editor

*italics*

_italics_

**bold**

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# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

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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.

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

The sum should start from \(0\).

Very true!

Seems interesting and damn tough!!!!

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

Ya I checked it .

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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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.

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

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The sum should start from \(0\).

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Very true!

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Seems interesting and damn tough!!!!

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

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Ya I checked it .

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