Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 14"
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<cmath>\begin{align*} S &= \frac{1}{8} + \frac{8}{64} + \frac{27}{512} + \frac{64}{4096} + \cdots \ | <cmath>\begin{align*} S &= \frac{1}{8} + \frac{8}{64} + \frac{27}{512} + \frac{64}{4096} + \cdots \ | ||
− | \frac{1}{8}S &= \quad \frac{1}{64} + \frac{8}{512} + \frac{27}{4096} + \cdots \ | + | \frac{1}{8}S &= \quad \; \; \; \frac{1}{64} + \frac{8}{512} + \frac{27}{4096} + \cdots \ |
\Rightarrow \frac{7}{8}S &= \frac{1}{8} + \frac{7}{64} + \frac{19}{512} + \frac{37}{4096} + \cdots \ | \Rightarrow \frac{7}{8}S &= \frac{1}{8} + \frac{7}{64} + \frac{19}{512} + \frac{37}{4096} + \cdots \ | ||
− | \frac{7}{64}S &= \quad \frac{1}{64} + \frac{7}{512} + \frac{19}{4096} + \cdots \ | + | \frac{7}{64}S &= \quad \; \; \; \frac{1}{64} + \frac{7}{512} + \frac{19}{4096} + \cdots \ |
\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 |
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+ | 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: | ||
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\frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{64 \cdot 7} \ | \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} \ |
Revision as of 17:21, 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:
\begin{align*} S &= \frac{1}{8} + \frac{8}{64} + \frac{27}{512} + \frac{64}{4096} + \cdots \\ \frac{1}{8}S &= \quad \; \; \; \frac{1}{64} + \frac{8}{512} + \frac{27}{4096} + \cdots \\ \Rightarrow \frac{7}{8}S &= \frac{1}{8} + \frac{7}{64} + \frac{19}{512} + \frac{37}{4096} + \cdots \\ \frac{7}{64}S &= \quad \; \; \; \frac{1}{64} + \frac{7}{512} + \frac{19}{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 \\ \Rightarrow \frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{512} + \frac{6}{4096} + \cdots 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} \\ \frac{343}{512}S &= \frac{97}{64 \cdot 7} \\ S &= \frac{776}{2401} \end{align*} (Error compiling LaTeX. Unknown error_msg)
Then our answer is .