# Identity using Choose numbers

• Jan 29th 2008, 06:30 PM
tbyou87
Identity using Choose numbers
Prove this identity:
$\sum_{k=0}^{n}$ ${n}\choose{k}$ = ${2n}\choose{n}$
Thanks
• Jan 29th 2008, 07:03 PM
mr fantastic
Quote:

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 ${2n}\choose{n}$ $= 2^n$ ......