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Math Help - Binomial coefficients

  1. #1
    MHF Contributor alexmahone's Avatar
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    Binomial coefficients

    Prove that for all positive integers n > 1,

    \sum_{k=0}^n\frac{1}{k+1}\dbinom{n}{k}(-1)^{k+1}=0

    My attempt:

    (1+x)^n=\sum_{k=0}^n\dbinom{n}{k}x^k

    Integrating both sides,

    \frac{(1+x)^{n+1}}{n+1}=\sum_{k=0}^n\dbinom{n}{k} \frac{x^{k+1}}{k+1}+C

    Substituting x = 0,

    C=\frac{1}{n+1}

    \frac{(1+x)^{n+1}}{n+1}=\sum_{k=0}^n\dbinom{n}{k} \frac{x^{k+1}}{k+1}+\frac{1}{n+1}

    Substituting x = -1,

    0=\sum_{k=0}^n\dbinom{n}{k} \frac{(-1)^{k+1}}{k+1}+\frac{1}{n+1}

    \sum_{k=0}^n\frac{1}{k+1}\dbinom{n}{k}(-1)^{k+1}=-\frac{1}{n+1}

    What am I doing wrong?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Binomial coefficients

    On second thoughts, I'm beginning to think that the book may be wrong and I may be right.
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  3. #3
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    Opalg's Avatar
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    Re: Binomial coefficients

    You are right and the question is wrong. You can check this by testing the formula with n=2. In that case, the sum becomes

    \sum_{k=0}^2\frac{1}{k+1}\dbinom{2}{k}(-1)^{k+1} = 1(-1) + \tfrac22(1) +\tfrac13(-1) = -\tfrac13.

    That agrees with your answer -\tfrac{1}{n+1} (with n=2), but not with the given answer 0.

    My guess is that the person who set this question forgot the constant when doing the integration. Congratulations to you for remembering that this is necessary.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Binomial coefficients

    Quote Originally Posted by chisigma View Post
    The correct procedure should be...

     (1+t)^{n} = \sum_{k=0}^{n} \binom{n}{k}\ t^{k} (1)

    ... then integrating both terms from 0 to x...

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

    ... then setting x=-1...

     \sum_{k=0}^{n} \binom{n}{k}\ \frac{(-1)^{k+1}}{k+1} =0 (3)

    Kind regards

    \chi \sigma
    After you integrated from 0 to x, you did not substitute for the lower limit 0.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: Binomial coefficients

    Quote Originally Posted by alexmahone View Post
    After you integrated from 0 to x, you did not substitute for the lower limit 0.
    I discovered that and I canceled my post before reading Your post...

    Kind regards

    \chi \sigma
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