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Thread: Summatory

  1. #1
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    Summatory

    Could you explain me how to get this result?

    Summatory-capturar.jpg
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  2. #2
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    Re: Summatory

    \sum_{j=1}^{n-1} (n-j)(n-j-k) = \sum_{k=1}^{n-1} k(k-2) = \sum_{k=1}^{n-1} (k^2 - 2k) = \sum_{k=1}^{n-1} k^2 - 2\sum_{k=1}^{n-1} k

    For the first step, think about the values that (n-j) takes. I've just replaced (n-j) with k and reversed the order of the terms in the sum.
    The remaining steps should be quite straightforward.
    The sums in the final expression are well known but easily derived:
    \sum_{i=1}^n \big(i^2 - (i-1)^2\big) = \sum_{i=1}^n \big(i^2 - (i^2-2i+1)\big) = \sum_{i=1}^n (2i-1)
    The left hand expression is a telescoping sum. You can use the equation to determine an expression for \sum_{i=1}^n i

    Similiarly:
    \sum_{i=1}^n \big(i^3 - (i-1)^3\big) = \sum_{i=1}^n \big(i^3 - (i^3-3i^2+3i-1)\big) = \sum_{i=1}^n (3i^2-3i+1)
    Which can be used to derive an expression for \sum_{i=1}^n i^2
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