Prove this identity:
$\displaystyle \sum_{k=0}^{n}$$\displaystyle {n}\choose{k}$ = $\displaystyle {2n}\choose{n}$
Thanks
By expanding $\displaystyle (1 + 1)^n$ using the binomial theorem it's simple to show that $\displaystyle \sum_{k=0}^{n}$$\displaystyle {n}\choose{k}$ = $\displaystyle 2^n$.
So you need to prove that $\displaystyle {2n}\choose{n}$ $\displaystyle = 2^n$ ......
Last edited by mr fantastic; Jan 29th 2008 at 06:08 PM.
Reason: Fixed up the latex