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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 $n= 1, 2, ...$ 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.

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

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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 $x_n = \beta^n$.

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