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Problem 242

AMC 10/12, early questions

Algebra Difficulty 3.8 Find the answer South African Mathematics Olympiad Third Round · South Africa

Two sequences of real numbers are defined as follows:
$u_1 = 0, \quad u_{n+1} = \frac{1}{2}(u_n + v_n)$
$v_1 = 1, \quad v_{n+1} = \frac{1}{4}(u_n + 3v_n)$
Find the value of $v_{2016} - u_{2016}$ .

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

$u_1 = 0$
$v_1 = 1$
$u_{n+1} = \frac{1}{2}(u_n + v_n)$
$v_{n+1} = \frac{1}{4}(u_n + 3v_n)$
$\begin{aligned} v_{2016} - u_{2016} &= \frac{1}{4}(u_{2015} + 3v_{2015}) - \frac{1}{2}(u_{2015} + v_{2015}) \\ &= \frac{1}{4}v_{2015} - \frac{1}{4}u_{2015} \\ &= \frac{1}{4}(v_{2015} - u_{2015}) \\ &= \frac{1}{4}\left(\frac{1}{4}(v_{2014} - u_{2014})\right) \\ &= \frac{1}{4}\left(\frac{1}{4}\left(\frac{1}{4}\right)^{2013}(v_1 - u_1)\right) \\ &= \frac{1}{4^{2015}}(v_1 - u_1) \\ &= \frac{1}{4^{2015}}(1 - 0) \\ &= \frac{1}{4^{2015}}. \end{aligned}$

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