Solving recurrence relations 4

Question

A sequence $a_n$ is defined by the recurrence relation:

$$a_n = 6a_{n−1} − 8a_{n-2}$$

for $2\le n\in\Bbb Z$.

Find all values of $r$ so that when $a_0 =1$ and $a_1 =r$, then $a_n =r^n$ for all $n\in\Bbb N$.

I am not sure where to begin.

Answer 1

HINT: Suppose that $a_n=r^n$ for all $n\in\Bbb N$. Then from the recurrence we must have

$$r^n=6r^{n-1}-8r^{n-2}\;.$$

Assuming that $r\ne 0$, we can divide through by $r^{n-2}$ to find that $r^2=6r-8$. You’ll have to check $r=0$ separately.