Let ${a_n}$ be a sequence whose corresponding power series $A(x)=\sum_{i\geq 0}a_ix^i$ satisfies
$$A(x)=\frac{6-x+5x^2}{1-3x^2-2x^3}$$
Determine a recurrence relation that ${a_n}$ satisfies.
I multiplied it out to
$$(1-3x^2-2x^3)A(x)=6-x+5x^2$$ $$a_0+(a_1-3a_0)x+(a_2-3a_1-2a_0)x^2+\dots = 6-x+5x^2$$ thus $$a_0=6$$ $$a_1=-1+3a_0$$ $$a_2=5+3a_1+2a_0$$ $$a_{n\geq 3}=3a_{n-1}+2a_{n-2}$$ but Wolfram Alpha instead gives the power series expansion as $$6-x+23 x^2+9 x^3+67 x^4+73 x^5+\cdots$$which does not satisfy my relation. Which is correct?
Hint: When multiplying $A(x)$ by the denominator, it seems you took $1-3x-2x^2$ instead of $1-3x^2-2x^3$.