Showing solutions for the recurrence relation $x_{n+1} = a x_n + b x_{n-1}$

Question

I am having trouble starting this question.

Given that the sequence $(x_n)$ satisfies the recurrence relation: $$x_{n+1} = a x_n + b x_{n-1},$$ where $n= 1, 2, \dots$ and $a$ and $b$ are given numbers, if the quadratic $r^2-ar-b = 0$ has distinct roots $\alpha$ and $\beta$, show that $x_n = \alpha^n$ and $x_n = \beta^n$ are both solutions of the recurrence relation shown above.

I have looked at this so many times and wherever I start I get nowhere, thanks for any help in advance.

Answer 1

As $\alpha$ is a solution of the mentioned equation, we have: $\alpha^2 = a\alpha + b$. Then, for $n \in \mathbb N$, we have $\alpha^{n-1} \times (\alpha^2) = \alpha^{n-1} \times (a \alpha + b)$, i.e. $\alpha^{n+1} = a \alpha^n + b\alpha^{n-1}$.

Answer 2

Substitute $x_n = \alpha^n$ into the recurrence relation to get

$$ \alpha^{n+1}=a\alpha^n+b\alpha^{n-1}.$$

Dividing by $\alpha^{n-1}$ and rearranging yields

$$ \alpha^{2}-a\alpha-b=0,$$

which is true since $\alpha$ is a root of the characteristic equation.

The same is true for $x_n = \beta^n$.