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Math Help - Finite Sum

  1. #1
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    Finite Sum

    Show that \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}-1}{n+1}.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Wizard2010 View Post
    Show that \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}-1}{n+1}.
    What have you tried? Have you hit it with induction yet?
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    What have you tried? Have you hit it with induction yet?
    I've actually realised that it's pretty easy. I've shown it (not by induction). Thanks.
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  4. #4
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    Quote Originally Posted by Wizard2010 View Post
    Show that \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}-1}{n+1}.
    \displaystyle (1+\alpha)^n = \sum_{k=0}^{n}\binom{n}{k}\alpha^k, therefore \displaystyle \int(1+\alpha)^n \;{d\alpha} = \int\sum_{k=0}^{n}\binom{n}{k}\alpha^k\;{d\alpha}.

    So \displaystyle \frac{(1+\alpha)^{n+1}}{n+1} + C = \sum_{k=0}^n \frac{\alpha^{k+1}}{k+1}\binom{n}{k}. Put \alpha = 0, to get C = -\frac{1}{n+1}.

    Thus \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}}{n+1}-\frac{1}{n+1} = \frac{(\alpha+1)^{n+1}-1}{n+1}.
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