Difference between revisions of "1969 Canadian MO Problems/Problem 6"
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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>. | 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 1== |
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> | 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. | ||
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In both cases, the expression telescopes into <math> (n+1)!-1.</math> | In both cases, the expression telescopes into <math> (n+1)!-1.</math> | ||
| + | == Solution 1== | ||
| + | We need to evaluate | ||
| + | <cmath>1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</cmath> | ||
| + | We replace <math>k\cdotk!</math> with <math>((k+1)-1)\cdotk!</math> | ||
| + | <cmath>(2-1)\cdot 1!+(3-1)\cdot 2!+(4-1)\cdot 3!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!</cmath> | ||
| + | Distribution yields | ||
| + | <cmath>(2\cdot 1!-1\cdot1!+3\cdot2!-1\cdot2!+\cdots+n(n-1)!-1(n-1)!+(n+1)n!-1\cdotn!</cmath> | ||
| + | Simplifying, | ||
| + | <cmath>2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!</cmath> | ||
| + | Which telescopes to | ||
| + | <cmath>(n+1)!-1!=\box((n+1)!-1)</cmath> | ||
{{Old CanadaMO box|num-b=5|num-a=7|year=1969}} | {{Old CanadaMO box|num-b=5|num-a=7|year=1969}} | ||
Revision as of 08:57, 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 1
We need to evaluate
![\[1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!\]](http://latex.artofproblemsolving.com/b/9/a/b9ad0e350c9af3f4f57820302c1653412c7a1cf5.png)
We replace $k\cdotk!$ (Error compiling LaTeX. Unknown error_msg) with $((k+1)-1)\cdotk!$ (Error compiling LaTeX. Unknown error_msg)
![\[(2-1)\cdot 1!+(3-1)\cdot 2!+(4-1)\cdot 3!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!\]](http://latex.artofproblemsolving.com/0/0/d/00de9f21f12c498752da542d7cf0013c3112f22d.png)
Distribution yields
\[(2\cdot 1!-1\cdot1!+3\cdot2!-1\cdot2!+\cdots+n(n-1)!-1(n-1)!+(n+1)n!-1\cdotn!\] (Error compiling LaTeX. Unknown error_msg)
Simplifying,
![\[2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!\]](http://latex.artofproblemsolving.com/b/2/b/b2beae3cc1eefe8ae56534fefb39167d582f02a1.png)
Which telescopes to
\[(n+1)!-1!=\box((n+1)!-1)\] (Error compiling LaTeX. Unknown error_msg)
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |
