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

Revision as of 11:36, 10 September 2008

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 5	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •	Followed by Problem 7

Revision as of 21:40, 17 November 2007 (view source) Temperal (talk \| contribs) (box) ← Older edit		Revision as of 11:36, 10 September 2008 (view source) 1=2 (talk \| contribs) (→‎Solution) Newer edit →
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−	{{Old CanadaMO box\|num-b=1\|num-a=3\|year=1969}}	+	{{Old CanadaMO box\|num-b=5\|num-a=7\|year=1969}}

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

Revision as of 11:36, 10 September 2008

Problem

Solution