Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 14"
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\Rightarrow \frac{49}{64}S &= \frac{1}{8} + \frac{6}{64} + \frac{12}{512} + \frac{18}{4096} + \cdots \ | \Rightarrow \frac{49}{64}S &= \frac{1}{8} + \frac{6}{64} + \frac{12}{512} + \frac{18}{4096} + \cdots \ | ||
\frac{49}{512}S &= \quad \; \; \; \frac{1}{64} + \frac{6}{512} + \frac{12}{4096} + \cdots \ | \frac{49}{512}S &= \quad \; \; \; \frac{1}{64} + \frac{6}{512} + \frac{12}{4096} + \cdots \ | ||
− | \Rightarrow \frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{512} + \frac{6}{4096} + \cdots | + | \Rightarrow \frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{512} + \frac{6}{4096} + \cdots \end{align*} </cmath> |
Almost done. Now that the numerators are constants instead of cubics, we can apply the formula for the sum of an infinite geometric series to get: | Almost done. Now that the numerators are constants instead of cubics, we can apply the formula for the sum of an infinite geometric series to get: | ||
− | \frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{64 \cdot 7} \ | + | <cmath>\begin{align*} \frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{64 \cdot 7} \ |
\frac{343}{512}S &= \frac{97}{64 \cdot 7} \ | \frac{343}{512}S &= \frac{97}{64 \cdot 7} \ | ||
S &= \frac{776}{2401} \end{align*} </cmath> | S &= \frac{776}{2401} \end{align*} </cmath> | ||
Then our answer is <math>\boxed{776}</math>. | Then our answer is <math>\boxed{776}</math>. |
Revision as of 17:22, 8 August 2021
Problem 14
Consider three infinite sequences of real numbers: It is known that, for all integers , the following statement holds: The elements of are defined by the relation . Let Then, can be represented as a fraction , where and are relatively prime positive integers. Find .
Solution
From the given condition, we have:
Then the sum becomes:
The final step is to take iterative differences, like so:
Almost done. Now that the numerators are constants instead of cubics, we can apply the formula for the sum of an infinite geometric series to get:
Then our answer is .