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Math Help - Simplifying summation

  1. #1
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    Simplifying summation

    Hello,

    I have a summation from k = 1,..., n of (m-k+1)(n-k+1), and I would like to simplify this in terms of just m and n.. I know that I need to write out the sum first, so

    sum = (mn) + (m-1)(n-1) + (m-2)(n-2) + ... + (m-n+1)(1),
    (is this right?)

    but I'm having trouble proceeding from this to the next step. Can anyone help?
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  2. #2
    Super Member

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    Hello, borracho!

    I would use these summation formulas:

    . . \sum^n_{k=1}k \;=\;\frac{k(k+1)}{2}

    . . \sum^n_{k=1}k^2 \:=\:\frac{k(k+1)(2k+1)}{6}


    S \;=\;\sum^m_{k=1}(m-k+1)(n-k+1)

    We have: . \bigg[(m+1)-k\bigg]\,\bigg[(n+1)-k\bigg] \;=\;(m+1)(n+1) - (m+n+2)k + k^2


    Then: . S \;\;=\;\;\sum^n_{k=1}\bigg[(m+1)(n+1) - (m+n+2)k + k^2\bigg]

    . . . . . S \;\;=\;\;(m+1)(n+1)\sum^n_{k=1}1 \;-\; (m+n+2)\sum^n_{k=1} k \;+\; \sum^n_{k=1}k^2

    . . . . . S \;\;=\;\;(m+1)(n+1)\cdot n \;-\;(m+n+2)\cdot\frac{n(n+1)}{2} \;+\;\frac{n(n+1)(2n+1)}{6}


    I'll let you simplify it . . .

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