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Math Help - Sum of C(n,r) proof

  1. #1
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    Tehran
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    Sum of C(n,r) proof

    Hi .
    Can Anyone help to proof \sum C(n,r)=2^n without using

    sets .

    I am Wondering how to Solve this Series?

    \sum_{r=0}^{n} \frac{n!}{r!(n-r)!}

    ............

    I Know :

    \sum_{r=0}^{n} \frac{n!}{r!(n-r)!} = n!\sum_{r=0}^{n} \frac{1}{r!(n-r)!}
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  2. #2
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    Quote Originally Posted by parkhid View Post
    Hi .
    Can Anyone help to proof \sum C(n,r)=2^n without using

    sets .

    I am Wondering how to Solve this Series?

    \sum_{r=0}^{n} \frac{n!}{r!(n-r)!}

    ............

    I Know :

    \sum_{r=0}^{n} \frac{n!}{r!(n-r)!} = n!\sum_{r=0}^{n} \frac{1}{r!(n-r)!}

    We know (I hope...) that (a,b)^n=\sum\limits_{k=0}^n \binom{n}{k} a^{n-k}b^k , so :

    \sum\limits_{k=0}^n \binom{n}{k}=\sum\limits_{k=0}^n \binom{n}{k}1^{n-k}\cdot 1^k= ...

    Tonio
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  3. #3
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    The Binomial Expansion is defined as:

    (x+a)^{n}=\sum_{k=0}^{n}{n \choose k}.x^{n-k}.a^{k} ,\forall x,a and n \in \mathbb{Z}_{+}

    Now, if x=a=1 , We have:

    2^{n}=\sum_{k=0}^{n}{n \choose k}

    What is the sum...
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  4. #4
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    Tehran
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    Question NO

    I will Know this way. i want to proof this by series and differential relations.
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  5. #5
    MHF Contributor
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    You can prove that \sum_{k=0}^n {n \choose k} = 2^n by induction on n.

    If n=1, then {1 \choose 0} + {1 \choose 1} = 2, so it is true for n=1.

    Now using the fact that {n \choose k} + {n \choose {k+1}} = {{n+1} \choose {k+1}}, and assuming that \sum_{k=0}^n {n \choose k} = 2^n, we have that

    \sum_{k=0}^{n+1} {{n+1} \choose k} = \sum_{k=1}^{n+1} {n \choose {k-1}} + \sum_{k=0}^n {n \choose k}

    \sum_{k=0}^{n+1} {{n+1} \choose k} = \sum_{k=0}^{n} {n \choose k} + \sum_{k=0}^n {n \choose k}

    \sum_{k=0}^{n+1} {{n+1} \choose k} = 2 \sum_{k=0}^{n} {n \choose k} = 2 \cdot 2^n = 2^{n+1}
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