Difference between revisions of "1969 Canadian MO Problems/Problem 6"
m (→Solution 1) |
m (→Solution 1) |
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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 | + | == Solution 2== |
− | + | ||
+ | We need to evaluate: | ||
<cmath>1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</cmath> | <cmath>1\cdot 1!+2\cdot 2!+3\cdot 3!+\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> |
Revision as of 10:01, 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:
We replace
with
Distribution yields
Simplifying,
Which telescopes to
So
is the solution.
1969 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |