Difference between revisions of "1969 Canadian MO Problems/Problem 6"
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<cmath>(n+1)!-1!=(n+1)!-1</cmath> | <cmath>(n+1)!-1!=(n+1)!-1</cmath> | ||
So <math>(n+1)!-1</math> is the solution. | 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 | ||
{{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 20:32, 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
1969 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |