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

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− | + | We notice that <math>x_5 + y_5 = 2x_4 + 2y_4 = 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))))).</math> Given that <math>x_0 = 3</math> and <math>y_0 = 1</math> in the problem, we can plug this in to get that <math>x_5 + y_5 = 2(2(2(2(2(3 + 1))))) = 2(2(2(2(2(4))))) = 128.</math> | |

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+ | We can use the same method to conclude that <math>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.</math> | ||

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+ | Adding this system of equations <math>x_5 + y_5 = 128</math> and <math>x_5 - y_5 = 486</math> gives us <math>2x_5 = 614.</math> Dividing both sides by <math>2</math>, results in <math>x_5 = 307.</math> | ||

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+ | ~mahaler |

## Revision as of 15:47, 11 July 2021

## Problem

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

Find .

## Solution

We notice that Given that and in the problem, we can plug this in to get that

We can use the same method to conclude that

Adding this system of equations and gives us Dividing both sides by , results in

~mahaler