Difference between revisions of "1969 Canadian MO Problems/Problem 6"
m (→Solution 1) |
m (→Solution 2) |
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| Line 14: | Line 14: | ||
We need to evaluate: | We need to evaluate: | ||
| − | <cmath>1\cdot 1!+2\cdot 2 | + | <cmath>1\cdot 1!+2\cdot 2!+\cdots+(n-1)(n-1)!+n\cdot n!</cmath> |
We replace <math>k\cdot k!</math> with <math>((k+1)-1)\cdot k!</math> | We replace <math>k\cdot k!</math> with <math>((k+1)-1)\cdot k!</math> | ||
| − | <cmath>(2-1)\cdot 1!+(3-1)\cdot 2 | + | <cmath>(2-1)\cdot 1!+(3-1)\cdot 2!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!</cmath> |
Distribution yields | Distribution yields | ||
<cmath>(2\cdot 1!-1\cdot1!+3\cdot2!-1\cdot2!+\cdots+n(n-1)!-1(n-1)!+(n+1)n!-1\cdot n!</cmath> | <cmath>(2\cdot 1!-1\cdot1!+3\cdot2!-1\cdot2!+\cdots+n(n-1)!-1(n-1)!+(n+1)n!-1\cdot n!</cmath> | ||
Revision as of 09:02, 3 December 2015
Problem
Find the sum of 
, where 
.
Solution 1
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 
Solution 2
We need to evaluate:
![\[1\cdot 1!+2\cdot 2!+\cdots+(n-1)(n-1)!+n\cdot n!\]](http://latex.artofproblemsolving.com/8/e/9/8e939967d601e87d0843bbb9bbe45a3ecc95341b.png)
We replace 
with 
![\[(2-1)\cdot 1!+(3-1)\cdot 2!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!\]](http://latex.artofproblemsolving.com/5/f/7/5f7f2ffbb0f03e85440220174016b6ea16ec0404.png)
Distribution yields
![\[(2\cdot 1!-1\cdot1!+3\cdot2!-1\cdot2!+\cdots+n(n-1)!-1(n-1)!+(n+1)n!-1\cdot n!\]](http://latex.artofproblemsolving.com/9/8/6/986a794be40d317825d5f56512768c170dfa11d5.png)
Simplifying,
![\[2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!\]](http://latex.artofproblemsolving.com/b/2/b/b2beae3cc1eefe8ae56534fefb39167d582f02a1.png)
Which telescopes to
![\[(n+1)!-1!=(n+1)!-1\]](http://latex.artofproblemsolving.com/5/c/3/5c321cf3a9620426b799582c86be54c46548d227.png)
So 
is the solution.
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |
