Difference between revisions of "1998 AJHSME Problems/Problem 12"
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Line 9: | Line 9: | ||
Taking the first product, we have | Taking the first product, we have | ||
− | <math>(1-\frac{1}{2})=\frac{1}{2}</math> | + | <math>\left(1-\frac{1}{2}\right)=\frac{1}{2}</math> |
<math>\frac{1}{2}\times2=1</math> | <math>\frac{1}{2}\times2=1</math> | ||
Line 36: | Line 36: | ||
<math>\frac{(9)(10)}{2}=45=\boxed{A}</math> | <math>\frac{(9)(10)}{2}=45=\boxed{A}</math> | ||
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== See also == | == See also == |
Revision as of 11:47, 20 December 2018
Problem
Solution
Taking the first product, we have
Looking at the second, we get
We seem to be going up by .
Just to check,
Now that we have discovered the pattern, we have to find the last term.
The sum of all numbers from to is
See also
1998 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.