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Math Help - Factoriel

  1. #1
    Super Member dhiab's Avatar
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    Factoriel : 1

    Prove :  \mathop \prod \limits_{i = 2}^n \left( {2i - 1} \right) = \frac{{\left( {2n} \right)!}}{{2^n \times n!}} <br />
    Last edited by dhiab; August 13th 2009 at 12:57 AM.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \prod_{i=2}^n(2i-1)=3\cdot5\cdot7\cdot\ldots\cdot(2n-1)=

    =\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot\ldots\cdot(2n)}{2\cdot 4\cdot 6\ldots\cdot(2n)}=\frac{(2n)!}{2^n(1\cdot 2\cdot 3\ldots\cdot n)}=\frac{(2n)!}{2^n\cdot n!}
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  3. #3
    Super Member Gamma's Avatar
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    Quote Originally Posted by dhiab View Post
    Prove :  \mathop \prod \limits_{i = 2}^n \left( {2i - 1} \right) = \frac{{\left( {2n} \right)!}}{{2^n \times n!}} <br />
    Induction, I trust you can do the base case.

    Assume  \mathop \prod \limits_{i = 2}^n \left( {2i - 1} \right) = \frac{{\left( {2n} \right)!}}{{2^n \times n!}}

    we must show  \mathop \prod \limits_{i = 2}^{n+1} \left( {2i - 1} \right) = \frac{{\left( {2(n+1)} \right)!}}{{2^{n+1} \times (n+1)!}}

     \mathop \prod \limits_{i = 2}^{n+1} \left( {2i - 1} \right) = \mathop (\prod \limits_{i = 2}^{n} \left( {2i - 1} \right) )(2(n+1)-1)=  \frac{2n!}{2^n\cdot n!}(2n+1)= \frac{2n!}{2^n\cdot n!}(2n+1)\left(\frac{2(n+1)}{2(n+1)}\right)= = \frac{{\left( {2(n+1)} \right)!}}{{2^{n+1} \times (n+1)!}}
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