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Thread: sums

  1. September 12th 2009, 06:43 AM #1
    thereddevils
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    sums

    Find the sum of

    <br /> \sum^n_{r=1}r(r!)<br />
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  2. September 12th 2009, 07:44 AM #2
    Defunkt
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    Try proving by induction, that:

    \sum^n_{r=1}r(r!) = (n+1)! - 1
    Last edited by Defunkt; September 12th 2009 at 12:04 PM.
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  3. September 12th 2009, 10:23 AM #3
    red_dog
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    \sum_{r=1}^nr\cdot r!=\sum_{r=1}^n(r+1-1)r!=\sum_{r=1}^n[(r+1)!-r!]=\sum_{r=1}^n(r+1)!-\sum_{r=1}^nr!=

    =2!+3!+\ldots+n!+(n+1)!-1!-2!-3!-\ldots-n!=(n+1)!-1
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