1969 Canadian MO Problems/Problem 6

Revision as of 15:20, 13 October 2006 by Chess64 (talk | contribs) (Solution)


Find the sum of $\displaystyle 1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!$, where $\displaystyle  n!=n(n-1)(n-2)\cdots2\cdot1$.


Note that for any positive integer $\displaystyle  n,$ $\displaystyle  n\cdot n!+(n-1)\cdot(n-1)!=(n^2+n-1)(n-1)!=(n+1)!-(n-1)!.$ Hence, pairing terms in the series will telescope most of the terms.

If $\displaystyle  n$ is odd, $\displaystyle  (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.$

If $\displaystyle  n$ is even, $\displaystyle  (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -3!+3!-1!.$ In both cases, the expression telescopes into $\displaystyle  (n+1)!-1.$

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