I am trying to solve the following problem:
Given the following recurrence relation: \begin{equation} a_n = 6a_{n-1} - 8a_{n-2} \ \text{ for } \ n \in \mathbb{Z}_{\ge2} \end{equation}
Find all values of $ r $ so that when $ a_0 = 1 $ and $ a_1 = r $ then $ a_n = r^n \ \ \forall n \in \mathbb{N} $
My attempt:
\begin{align} \text{By distinct roots theorem,} \\ \\ \text{Characteristic equation: } & t^2 - 6t + 8 = 0 \\ & (t-3)^2 + 8 - 9 = 0 \\ & (t-3)^2 = 1 \\ & t = 4 \text{ or } t = 2 \end{align} \begin{align} \text{Let } a_0 = 1 = C(4^0) + D(2^0) \text{ where C, D } \in \mathbb{R} \end{align} \begin{align} 1 = C + D \\ C = 1 - D \end{align}
\begin{align} \text{Let } a_1 = r &= C(4) + D(2) \\ r &= 4C + 2D \\ r &= 4(1-D) + 2D \quad \text{(By substitution)} \\ r &= 4 - 4D + 2D \\ r &= 4 - 2D \end{align}
It seems that I must have gone wrong somewhere as $ r < 0 $ for $ D > 2$. Hence the answer for $ a_n $ would be very strange (i.e. $ a_n < 0 $ when $ r = -4 $). Could anyone please assist me?
By substituting $a_n$ by $r^n$, you get $r^n = 6 r^{n-1} -8r^{n-2}$ for $n\ge2$.
Thus $r^2 - 6r + 8 =0$ hence $r \in \{2, 4\}$ as you rightly figured out.
The issue with what you did is that you tried to solve the equations once you had $a_n = C 4^n + D 2^n$. You should have figured that either $C = 0$ and $D = 1$ or $D = 0$ and $C = 1$ (or else there exists no $r \in \Bbb R$ such that $\forall n\in \Bbb N, a_n = r^n$)