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> \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> | ||
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<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> | <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> |
Revision as of 19:14, 22 July 2006
Problem
Solution
Factoring out a telescopes the sum to .