Find the solution for an = 6a,-1-8an-2 for n > 2 and ao = 4, a1 = 10.

Given recurrence relation is an=6an-1-8an-2, n≥2 and a0=4, a1=10. Let an=xn be a solution of above…

Answered: Find the solution for an = 6a-1-8an-2…

PREALGEBRA

15th Edition

ISBN:9781938168994

Author:OpenStax

Publisher:OpenStax

Chapter11: Graphs

Section11.1: Use The Rectangular Coordinate System

Problem 11.28TI: Find three solutions to the equation: x + 5y = 10.

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Find the solution for an = 6a,-1-8an-2 for n > 2 and ao = 4, a1 = 10.

Expert Solution

Step 1

Given recurrence relation is $a_{n} = 6 a_{n - 1} - 8 a_{n - 2}, n \geq 2$ and $a_{0} = 4, a_{1} = 10$ .

Let $a_{n} = x^{n}$ be a solution of above recurrence relation. Therefore above recurrence relation can be solved as:

$a_{n} = 6 a_{n - 1} - 8 a_{n - 2} x^{n} = 6 x^{n - 1} - 8 x^{n - 2}$ , Hence

$\begin{array}{rcl} x^{n} - 6 x^{n - 1} + 8 x^{n - 2} & = & 0 \\ x^{n - 2} (x^{2} - 6 x + 8) & = & 0 \\ x^{2} - 6 x + 8 & = & 0 \\ (x - 2) (x - 4) & = & 0 \end{array}$

Therefore either $x - 2 = 0$ or $x - 4 = 0$ . Hence either $x = 2$ or $x = 4$ .

Therefore $a_{n} = 2^{n}$ or $a_{n} = 4^{n}$ .

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