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Difference between revisions of "1969 Canadian MO Problems/Problem 6"

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In both cases, the expression telescopes into <math>\displaystyle  (n+1)!-1.</math>
 
In both cases, the expression telescopes into <math>\displaystyle  (n+1)!-1.</math>
  
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* [[1969 Canadian MO Problems/Problem 5|Previous Problem]]
 
* [[1969 Canadian MO Problems/Problem 5|Previous Problem]]
 
* [[1969 Canadian MO Problems/Problem 7|Next Problem]]
 
* [[1969 Canadian MO Problems/Problem 7|Next Problem]]
 
* [[1969 Canadian MO Problems|Back to Exam]]
 
* [[1969 Canadian MO Problems|Back to Exam]]

Revision as of 16:08, 12 October 2006

Problem

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

Solution

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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