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Math Help - Summation Closed Forms

  1. #1
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    Summation Closed Forms

    Find the closed form for:
    \sum_{i=-1}^{n}i

    It's given that the closed form \sum_{i=1}^{n}i = (n(n+1))/2 so we can write:
    \sum_{i=-1}^{n}i = (-1) + 0 + \sum_{i=1}^{n}i = (n(n+1))/2

    ...

    ( (n+2)(n-1) )/2

    I understand the final closed form answer but could someone explain where the terms (-1) + 0 come from on the first line. I'm assuming the -1 comes from the lower bounds of the summation..
    Last edited by mr fantastic; February 3rd 2009 at 12:24 PM. Reason: Deleted question restored
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  2. #2
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    \sum\limits_{i=-1}^{n}{i}=\sum\limits_{i=1}^{n+2}{(i-2)}=\frac{(n+2)(n+3)}{2}-2(n+2)=\frac{(n+2)(n-1)}{2}.
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  3. #3
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    Thanks.
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