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 \end{align*} </cmath> | \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 <math>(6)</math> instead of cubics <math>(n^3)</math>, we can apply the formula for the sum of an infinite geometric series to get: |
<cmath>\begin{align*} \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} \ |
Latest revision as of 17:23, 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 .