Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 30"
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Factoring out a <math>\frac{1}{3}</math> telescopes the sum to <math>\frac{1}{3}\left(\frac{1}{1 \cdot 2 \cdot 3}-\frac{1}{29 \cdot 30 \cdot 31}\right) = \frac{749}{13485}</math>. | Factoring out a <math>\frac{1}{3}</math> telescopes the sum to <math>\frac{1}{3}\left(\frac{1}{1 \cdot 2 \cdot 3}-\frac{1}{29 \cdot 30 \cdot 31}\right) = \frac{749}{13485}</math>. | ||
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− | * [[University of South Carolina High School Math Contest/1993 Exam]] | + | *[[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Previous Problem]] |
+ | *[[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]] | ||
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+ | [[Category:Intermediate Algebra Problems]] |