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Thread: Binomial coefficients 3

  1. #1
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    Binomial coefficients 3

    This problem is very difficult...can someone help me with this please? ...this is all about binomials



    1. Show that


    $\displaystyle
    \sum_{k=1}^{n}k\binom{n}{2k+1}=(n-2){2}^{n-3}
    $
    Last edited by manusform; Sep 16th 2009 at 11:26 PM. Reason: equation is not appropriate
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  2. #2
    Member
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    Hello,
    $\displaystyle (1-x)^n=\sum_{k=0}^n (-1)^k \binom{n}{k}x^k=-\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1} + \sum_{k=0}^n \binom{n}{2k}x^{2k}$

    $\displaystyle (1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k=\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1 } + \sum_{k=0}^n \binom{n}{2k}x^{2k}$

    Take derivative with respect to $\displaystyle x$ of both equalities

    $\displaystyle -n(1-x)^{n-1}=-\sum_{k=0}^n (2k+1) \binom{n}{2k+1}x^{2k} + \sum_{k=0}^n 2k \binom{n}{2k}x^{2k-1}$

    $\displaystyle n(1+x)^{n-1}=\sum_{k=0}^n (2k+1) \binom{n}{2k+1}x^{2k} + \sum_{k=0}^n 2k \binom{n}{2k}x^{2k-1}$

    Set $\displaystyle x=1$ in all four equalities and you get

    $\displaystyle \sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} =2^{n-1}$

    $\displaystyle \sum_{k=0}^n (2k+1) \binom{n}{2k+1} = \sum_{k=0}^n 2k \binom{n}{2k} = n2^{n-2}$ for $\displaystyle n>1$

    Now because

    $\displaystyle \sum_{k=0}^n (2k+1) \binom{n}{2k+1} = 2\sum_{k=0}^n k \binom{n}{2k+1}+\sum_{k=0}^n \binom{n}{2k+1}$

    you get

    $\displaystyle \sum_{k=0}^n k \binom{n}{2k+1} = \frac{1}{2}(n2^{n-2} - 2^{n-1}) =\frac{1}{2}(n2^{n-2} - 2\cdot2^{n-2})=(n-2)2^{n-3}$ for $\displaystyle n>1$
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  3. #3
    Junior Member
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    I think you made a mistake.

    $\displaystyle \sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} \neq2^{n-1}$

    The right form is:

    $\displaystyle \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2k +1} = \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \binom{n}{2k} =2^{n-1}$
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  4. #4
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    Sorry, both two form are right.
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  5. #5
    Member
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    Hi streethot,

    I don't think it is a mistake, since the definition of binomial coefficients says
    $\displaystyle \binom{n}{k}=0 $ for $\displaystyle k>n$. This means your form and my form are equivalent.

    I gave a derivation of $\displaystyle \sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} = 2^{n-1}$ above.
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