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Thread: SUM

  1. #1
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    SUM

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  2. #2
    Junior Member Bradyns's Avatar
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    Re: SUM

    Does this help?
    $\displaystyle \sum_{n=1}^{20}n(21-n)$

    sum n(21-n) from n=1 through 20 - Wolfram|Alpha
    Thanks from abualabed
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  3. #3
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    Re: SUM

    Quote Originally Posted by abualabed View Post

    You can use the special sum $\displaystyle \sum\limits_{k = 1}^n {{k^2}} = \frac{{n(n + 1(2n + 1)}}{6}$.

    Write your sum as $\displaystyle \sum\limits_{k = 1}^{20} {k(21 - k)} = 21\sum\limits_{k = 1}^{20} k - \sum\limits_{k = 1}^{20} {{k^2}}~.$
    Thanks from MarkFL and abualabed
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