Difference between revisions of "1969 Canadian MO Problems/Problem 6"
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== Problem == | == Problem == | ||
| − | Find the sum of <math> | + | 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> | + | 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> | + | If <math> n</math> is [[odd integer | odd]], <math> (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.</math> |
| − | If <math> | + | 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> | + | In both cases, the expression telescopes into <math> (n+1)!-1.</math> |
| − | -- | + | {{Old CanadaMO box|num-b=1|num-a=3|year=1969}} |
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Revision as of 22:40, 17 November 2007
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 
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 3 |
