Identity using Choose numbers

**tbyou87** · January 29th 2008, 05:30 PM

Prove this identity:
$\sum_{k=0}^{n}$ ${n}\choose{k}$ = ${2n}\choose{n}$
Thanks

**mr fantastic** · January 29th 2008, 06:03 PM

Originally Posted by tbyou87

Prove this identity:
$\sum_{k=0}^{n}$ ${n}\choose{k}$ = ${2n}\choose{n}$
Thanks

By expanding $(1 + 1)^n$ using the binomial theorem it's simple to show that $\sum_{k=0}^{n}$ ${n}\choose{k}$ = $2^n$ .

So you need to prove that ${2n}\choose{n}$ $= 2^n$ ......

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