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University of South Carolina High School Math Contest/1993 Exam/Problem 30

Revision as of 14:47, 19 August 2006 by JBL (talk | contribs) (→‎Solution)

Problem

$\frac 1{1\cdot 2\cdot 3\cdot 4} + \frac 1{2\cdot 3\cdot 4\cdot 5} + \frac 1{3\cdot 4\cdot 5\cdot 6} + \cdots + \frac 1{28\cdot 29\cdot 30\cdot 31} =$


$\mathrm{(A) \ }1/18 \qquad \mathrm{(B) \ }1/21 \qquad \mathrm{(C) \ }4/93 \qquad \mathrm{(D) \ }128/2505 \qquad \mathrm{(E) \ } 749/13485$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Factoring out a $\frac{1}{3}$ telescopes the sum to $\frac{1}{3}\left(\frac{1}{1 \cdot 2 \cdot 3}-\frac{1}{29 \cdot 30 \cdot 31}\right) = \frac{749}{13485}$.

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