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Thread: Identity using Choose numbers

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    Identity using Choose numbers

    Prove this identity:
    $\displaystyle \sum_{k=0}^{n}$$\displaystyle {n}\choose{k}$ = $\displaystyle {2n}\choose{n}$
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    Quote Originally Posted by tbyou87 View Post
    Prove this identity:
    $\displaystyle \sum_{k=0}^{n}$$\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
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