Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 30"
(→Solution) |
|||
(5 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
== 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> | ||
+ | |||
<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> | ||
Line 7: | Line 8: | ||
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>. | ||
− | + | ---- | |
− | * [[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]] | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |