2021 JMPSC Invitationals Problems/Problem 4

Problem

Let $(x_n)_{n\geq 0}$ and $(y_n)_{n\geq 0}$ be sequences of real numbers such that $x_0 = 3$ , $y_0 = 1$ , and, for all positive integers $n$ ,

$x_{n+1}+y_{n+1} = 2x_n + 2y_n,$ $x_{n+1}-y_{n+1}=3x_n-3y_n.$ Find $x_5$ .

Solution

We notice that $x_5 + y_5 = 2(x_4 + y_4)$ $= 2(2(x_3 + y_3))$ $= 2(2(2(x_2 + y_2)))$ $= 2(2(2(2(x_1 + y_1))))$ $= 2(2(2(2(2(x_0 + y_0))))).$ Since we are given that $x_0 = 3$ and $y_0 = 1$ , we can plug these values in to get that $x_5 + y_5 = 2(2(2(2(2(3 + 1))))) = 2(2(2(2(2(4))))) = 128 \qquad (1).$