sums

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September 12th 2009, 06:43 AM
thereddevils

sums

Find the sum of

$<br /> \sum^n_{r=1}r(r!)<br />$
September 12th 2009, 07:44 AM
Defunkt

Try proving by induction, that:

$\sum^n_{r=1}r(r!) = (n+1)! - 1$
September 12th 2009, 10:23 AM
red_dog

$\sum_{r=1}^nr\cdot r!=\sum_{r=1}^n(r+1-1)r!=\sum_{r=1}^n[(r+1)!-r!]=\sum_{r=1}^n(r+1)!-\sum_{r=1}^nr!=$

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

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