Difference between revisions of "1969 Canadian MO Problems/Problem 6"
(→Solution) |
m (→Solution 1) |
||
Line 14: | Line 14: | ||
We need to evaluate | 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\ | + | We replace <math>k\cdot k!</math> with <math>((k+1)-1)\cdot k!</math> |
<cmath>(2-1)\cdot 1!+(3-1)\cdot 2!+(4-1)\cdot 3!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!</cmath> | <cmath>(2-1)\cdot 1!+(3-1)\cdot 2!+(4-1)\cdot 3!+\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\ | + | <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, | Simplifying, | ||
<cmath>2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!</cmath> | <cmath>2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!</cmath> |
Revision as of 08:59, 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 1
We need to evaluate
We replace with Distribution yields Simplifying, Which telescopes to
\[(n+1)!-1!=\box((n+1)!-1)\] (Error compiling LaTeX. Unknown error_msg)
1969 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |