Difference between revisions of "1969 Canadian MO Problems/Problem 6"
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== 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> | + | 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> |
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, <math>\displaystyle (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.</math> | + | If <math>\displaystyle n</math> is [[odd integer | odd]], <math>\displaystyle (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.</math> |
| − | If <math>\displaystyle n</math> is even, <math>\displaystyle (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -3!+3!-1!.</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> |
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> | ||
Revision as of 17:07, 12 October 2006
Problem
Find the sum of 
, where 
.
Solution
Note that for any positive integer 
Hence, pairing terms in the series will telescope most of the terms.
If 
is odd, 
If is even, 
In both cases, the expression telescopes into 
