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Consider a Fibonacci type of sequence

$$a_0=1~, \qquad a_1=2~, \qquad 3a_{n+1}=a_n+2a_{n-1}~, \quad n=1,2, \ldots$$

Find the formula for the general nth term of the sequence.

I'm having trouble starting this one, I do have another example that goes :

$$F_1=2~, \quad F_2=4~, \quad F_{n+1}=3F_n+5F_{n-1} ~, \quad n=2,3\ldots$$

With the example I know that $$F_n=\alpha r^n$$ Then $$\alpha r^{n+1}=3\alpha r^n+5\alpha r^{n-1}$$

Now the only help I need for now is the initial step would it be

$$3 \alpha r^{n+1}= \alpha r^n+2\alpha r^{n-1}$$

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Hint: Your last line implies $3r^2 = r + 2$ because $\alpha r\ne0$ (why?). Solve for $r$.

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