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

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==Problem== | ==Problem== | ||

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

− | + | ||

− | x_{n+1}+y_{n+1} | + | <cmath>x_{n+1}+y_{n+1} = 2x_n + 2y_n,</cmath> |

− | x_{n+1}-y_{n+1} | + | <cmath>x_{n+1}-y_{n+1}=3x_n-3y_n.</cmath> |

− | + | Find <math>x_5</math>. | |

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

− | + | 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> | ||

+ | |||

+ | Adding <math>(1)</math> and <math>(2)</math> gives us <math>2 \cdot x_5 = 614.</math> Dividing both sides by <math>2</math> yields <math>x_5 = \boxed{307}.</math> | ||

+ | |||

+ | ~mahaler | ||

+ | |||

+ | ==Solution 2== | ||

+ | Add both equations to get <math>2(x_{n+1})=5x_n-y_n</math>, and subtract both equations to get <math>2(y_{n+1})=5y_n-x_n</math>, so now we bash: <math>x_1=7</math> and <math>y_1=1</math>. <math>x_2=17</math> and <math>y_2=-1</math>. <math>x_3=43</math> and <math>y_3=-11</math>. <math>x_4=113</math> and <math>y_4=-49</math>, <math>x_5=\frac{614}{2}=\boxed{307}</math> | ||

+ | |||

+ | ~Geometry285 | ||

+ | |||

+ | ==See also== | ||

+ | #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]] | ||

+ | #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]] | ||

+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||

+ | {{JMPSC Notice}} |

## Latest revision as of 21:08, 11 July 2021

## Contents

## 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

## Solution 2

Add both equations to get , and subtract both equations to get , so now we bash: and . and . and . and ,

~Geometry285

## See also

- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.