# Northeastern WOOTers Mock AIME I Problems/Problem 14

## Problem 14

Consider three infinite sequences of real numbers: $\begin{eqnarray*} X &=& \left( x_1, x_2, \cdots \right), \\ Y &=& \left( y_1, y_2, \cdots \right), \\ Z &=& \left( z_1, z_2, \cdots \right). \end{eqnarray*}$ It is known that, for all integers $n$, the following statement holds: \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.\end{align*}The elements of $Y$ are defined by the relation $y_n=2^{\frac{n}{2^n}}$. Let $$S =\sum_{n=1}^{\infty} \log_2 x_n \log_2 y_n \log_2 z_n.$$Then, $S$ can be represented as a fraction $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m$.

## Solution

From the given condition, we have:

\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) \\ \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*}

Then the sum becomes:

\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*}

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 \end{align*}

Almost done. Now that the numerators are constants $(6)$ instead of cubics $(n^3)$, we can apply the formula for the sum of an infinite geometric series to get:

\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*}

Then our answer is $\boxed{776}$.