# Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 4"

(→Solution) |
(→Solution) |
||

Line 7: | Line 7: | ||

==Solution== | ==Solution== | ||

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

Similarly, we conclude that <cmath>x_5 - y_5 = 3(3(3(3(3(x_0 - y_0)))))= 3(3(3(3(3(3-1))))) = 3(3(3(3(3(2))))) = 486 \qquad (2).</cmath> | Similarly, we conclude that <cmath>x_5 - y_5 = 3(3(3(3(3(x_0 - y_0)))))= 3(3(3(3(3(3-1))))) = 3(3(3(3(3(2))))) = 486 \qquad (2).</cmath> |

## Revision as of 15:54, 11 July 2021

## Problem

Let and be sequences of real numbers such that , , and, for all positive integers ,

Find .

## Solution

We notice that Since we are given that and , we can plug these values in to get that

Similarly, we conclude that

Adding and gives us Dividing both sides by yields

~mahaler