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. | ||
| Line 11: | Line 11: | ||
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 2== | ||
| + | |||
| + | We need to evaluate: | ||
| + | <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> | ||
| + | <cmath>(2-1)\cdot 1!+(3-1)\cdot 2!+\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\cdot n!</cmath> | ||
| + | Simplifying, | ||
| + | <cmath>2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!</cmath> | ||
| + | Which telescopes to | ||
| + | <cmath>(n+1)!-1!=(n+1)!-1</cmath> | ||
| + | So <math>(n+1)!-1</math> is the solution. | ||
| + | |||
| + | == Solution 3== | ||
| + | |||
| + | <math>S=\sum_{k=1}^{n}k\cdot k!=\sum_{k=1}^{n}(k\cdot k!+k!-k!)=\sum_{k=1}^{n}\left( (k+1)\cdot k!-k! \right)=\sum_{k=1}^{n}(k+1)\cdot k!-\sum_{k=1}^{n}k!\\ | ||
| + | =\sum_{k=2}^{n+1}k!-\sum_{k=1}^{n}k!=(n+1)!+\sum_{k=2}^{n}k!-\sum_{k=2}^{n}k!-1!=\boxed{(n+1)!-1}</math> | ||
| + | |||
| + | ~Tomas Diaz. orders@tomasdiaz.com | ||
| + | |||
| + | {{alternate solutions}} | ||
{{Old CanadaMO box|num-b=5|num-a=7|year=1969}} | {{Old CanadaMO box|num-b=5|num-a=7|year=1969}} | ||
Latest revision as of 19:33, 27 November 2023
Contents
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.
Solution 3
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |
