dangerousliri 05.03.2013 14:51 Let be $n$ positive integer than calculate: $1\cdot 1!+2\cdot2!+...+n\cdot n!$
97aydos 05.03.2013 15:05 $ 1\cdot1!+2\cdot2!+...+n \cdot n!= (2-1)\cdot1!+(3-1)\cdot2!+...+(n-1)\cdot n!= (2!-1!)+(3!-2!)+...+((n+1)!-n!)=(n+1)!-1$
ssilwa 06.03.2013 01:06 @danger: If you didnt see that approach right off the bat, ( I sure didnt!) then you can guess a few terms and try induction which works perfectly.