Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 14"
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<cmath>\begin{align*} \left( \left( \log_2 x_n \right)^2 + \left( \log_2 y_n \right)^2 \right) \cdot \left( \left( \log_2 y_n \right)^2 + \left( \log_2 z_n \right)^2 \right) &= \left( \log_2 x_n \log_2 y_n + \log_2 y_n \log_2 z_n \right)^2 \\ | <cmath>\begin{align*} \left( \left( \log_2 x_n \right)^2 + \left( \log_2 y_n \right)^2 \right) \cdot \left( \left( \log_2 y_n \right)^2 + \left( \log_2 z_n \right)^2 \right) &= \left( \log_2 x_n \log_2 y_n + \log_2 y_n \log_2 z_n \right)^2 \\ | ||
− | \left( \log_2 y_n \right)^4 + \left( \log_2 x_n \log_2 z_n \right)^2 &= 2 \left( \log_2 y_n \right)^2 \left( \log_2 x_n \log_2 z_n \right) \end{align*} </cmath> | + | \left( \log_2 y_n \right)^4 + \left( \log_2 x_n \log_2 z_n \right)^2 &= 2 \left( \log_2 y_n \right)^2 \left( \log_2 x_n \log_2 z_n \right) \\ |
+ | \left( \left( \log_2 y_n \right)^2 - \log_2 x_n \log_2 z_n \right)^2 &= 0 \\ | ||
+ | \left( \log_2 y_n \right)^2 &= \log_2 x_n \log_2 z_n \end{align*} </cmath> | ||
+ | |||
+ | Then the sum becomes: | ||
+ | |||
+ | <cmath>\begin{align*} S &= \sum_{n=1}^{\infty} \log_2 x_n \log_2 y_n \log_2 z_n \\ | ||
+ | S &= \sum_{n=1}^{\infty} \frac{n^3}{8^n} \end{align*} </cmath> | ||
+ | |||
+ | The final step is to take iterative differences, like so: | ||
+ | |||
+ | <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 \\ | ||
+ | \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 \end{align*} </cmath> | ||
+ | |||
+ | 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} \\ | ||
+ | \frac{343}{512}S &= \frac{97}{64 \cdot 7} \\ | ||
+ | S &= \frac{776}{2401} \end{align*} </cmath> | ||
+ | |||
+ | Then our answer is <math>\boxed{776}</math>. |
Latest revision as of 18:23, 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 .