Difference between revisions of "1969 Canadian MO Problems/Problem 6"
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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 | + | == Solution 2== |
− | + | ||
− | <cmath>1\cdot 1!+2\cdot 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> | We replace <math>k\cdot k!</math> with <math>((k+1)-1)\cdot k!</math> | ||
− | <cmath>(2-1)\cdot 1!+(3-1)\cdot 2 | + | <cmath>(2-1)\cdot 1!+(3-1)\cdot 2!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!</cmath> |
Distribution yields | 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> | <cmath>(2\cdot 1!-1\cdot1!+3\cdot2!-1\cdot2!+\cdots+n(n-1)!-1(n-1)!+(n+1)n!-1\cdot n!</cmath> |
Revision as of 10:02, 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 |