Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 30"
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== Problem == | == Problem == | ||
| + | <center><math> \frac 1{1\cdot 2\cdot 3\cdot 4} + \frac 1{2\cdot 3\cdot 4\cdot 5} + \frac 1{3\cdot 4\cdot 5\cdot 6} + \cdots + \frac 1{28\cdot 29\cdot 30\cdot 31} = </math></center> | ||
| − | <center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ } </math></center> | + | <center><math> \mathrm{(A) \ }1/18 \qquad \mathrm{(B) \ }1/21 \qquad \mathrm{(C) \ }4/93 \qquad \mathrm{(D) \ }128/2505 \qquad \mathrm{(E) \ } 749/13485</math></center> |
== Solution == | == Solution == | ||
| + | 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>. | ||
== See also == | == See also == | ||
* [[University of South Carolina High School Math Contest/1993 Exam]] | * [[University of South Carolina High School Math Contest/1993 Exam]] | ||
Revision as of 19:13, 22 July 2006
Problem
Solution
Factoring out a 
telescopes the sum to 
.
