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 1== |
− | 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> |
+ | == 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}} |
Latest revision as of 20:33, 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:
We replace
with
Distribution yields
Simplifying,
Which telescopes to
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 |