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Sum

Posted: 06/04-2011 12:49
by Gustav
Beregn

[tex]\sum_{k=1}^{n} k\cdot k![/tex]

Posted: 10/04-2011 15:22
by Karl_Erik
[tex]\sum_{k=1}^n k \cdot k! = \sum_{k=1}^n (k+1) \cdot k! - k! = \sum_{k=1}^n (k+1)!- k! = (n+1)!-1!=(n+1)!-1[/tex]