I'm trying to figure out how to convert this recursive formula $A_n = \frac{8}{9} A_{n-1} + \frac{6}{5} \left(\frac{20}{9}\right)^n $ into $A_n = 4 \left(\frac{8}{9}\right)^n + 2 \left(\frac{20}{9}\right)^n$ and I'm not sure how to approach it.
The first thing I thought of was to try to find a pattern by substituting a previous term into the next one \begin{align*} A_n & = \frac{8}{9} A_{n-1} + \frac{6}{5} \left(\frac{20}{9}\right)^n \\ A_{n+1} & = \frac{8}{9} \left( \frac{8}{9} A_{n-1} + \frac{6}{5} \left(\frac{20}{9}\right)^n\right) + \frac{6}{5} \left(\frac{20}{9}\right)^{n+1} \\ & = \left(\frac{8}{9}\right)^2 A_{n-1}+ \frac{6}{5}\frac{8}{9}\left(\frac{20}{9}\right)^n + \frac{6}{5} \left(\frac{20}{9}\right)^{n+1} \\ & = \left(\frac{8}{9}\right)^2 A_{n-1}+ \frac{6}{5}\left(\frac{20}{9}\right)^n \left(\frac{8}{9}+\frac{20}{9}\right) \end{align*} But that didn't seem to work, although it's possible I simplified incorrectly (or not in the way you're supposed to)
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Actually, I think I forgot a crucial piece of information: $A_0 = 6$
$$A_{n+1}=aA_n+bc^n.$$
Try a solution of the form $dc^n$. You have
$$dc^{n+1}=adc^n+bc^n,$$ giving $$d=\frac b{c-a}.$$
Now by subtraction,
$$A_{n+1}-dc^{n+1}=aA_n+bc^n-dc^{n+1}$$
is $$(A_{n+1}-dc^{n+1})=a(A_n-dc^n),$$
which is easy to solve for
$$A_n-dc^n$$ as it is a geometric progression.