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

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== Problem ==
 
== Problem ==
Find the sum of <math>\displaystyle 1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</math>, where <math>\displaystyle  n!=n(n-1)(n-2)\cdots2\cdot1</math>.
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Find the sum of <math>1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</math>, where <math> n!=n(n-1)(n-2)\cdots2\cdot1</math>.
  
 
== Solution ==
 
== Solution ==
Note that for any [[positive integer]] <math>\displaystyle  n,</math> <math>\displaystyle  n\cdot n!+(n-1)\cdot(n-1)!=(n^2+n-1)(n-1)!=(n+1)!-(n-1)!.</math>
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Note that for any [[positive integer]] <math> n,</math> <math> n\cdot n!+(n-1)\cdot(n-1)!=(n^2+n-1)(n-1)!=(n+1)!-(n-1)!.</math>
 
Hence, pairing terms in the series will telescope most of the terms.
 
Hence, pairing terms in the series will telescope most of the terms.
  
If <math>\displaystyle  n</math> is [[odd integer | odd]], <math>\displaystyle  (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.</math>
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If <math> n</math> is [[odd integer | odd]], <math> (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.</math>
  
If <math>\displaystyle  n</math> is [[even integer | even]], <math>\displaystyle  (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -3!+3!-1!.</math>
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If <math> n</math> is [[even integer | even]], <math> (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -3!+3!-1!.</math>
In both cases, the expression telescopes into <math>\displaystyle  (n+1)!-1.</math>
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In both cases, the expression telescopes into <math> (n+1)!-1.</math>
  
  
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{{Old CanadaMO box|num-b=1|num-a=3|year=1969}}
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* [[1969 Canadian MO Problems/Problem 7|Next Problem]]
 
* [[1969 Canadian MO Problems|Back to Exam]]
 

Revision as of 22:40, 17 November 2007

Problem

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

Solution

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

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

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


1969 Canadian MO (Problems)
Preceded by
Problem 1
1 2 3 4 5 6 7 8 Followed by
Problem 3


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