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Math Help - combinatorial sum

  1. #1
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    combinatorial sum

    \sum_{n=2}^x {1\over{x}} (1+{1\over{x}})^{n-2}

    anyone have a way of making this prettier?
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  2. #2
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    Quote Originally Posted by jbpellerin View Post
    \sum_{n=2}^x {1\over{x}} (1+{1\over{x}})^{n-2}

    anyone have a way of making this prettier?
    \sum_{n=2}^x {1\over{x}} (1+{1\over{x}})^{n-2} is equivalent to


    \sum_{n=2}^x  \frac {1}{x}(\frac{x+1}{x})^{n-2}

    \sum_{n=2}^x \frac{(x+1)^{n-2}}{{x}^{n-1}}

    Can you take it from there?
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  3. #3
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    actually it turned out to be pretty easy
    (1+{1\over{x}})^{x-1} -1
    it's a geometric series
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  4. #4
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    First observation : you can factor out 1/x from the sum :
    =\frac 1x \sum_{n=2}^x \left(1+\frac 1x\right)^{n-2}

    Second observation : what you said > geometric series !
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