Why the generating function of $x^2(1-3x)^{-1}$ is $(0,0,1,3,3^2,3^3)$ instead of $(0,0,1,-3,3^2,-3^3)$?

Question

I've made this exercise. First I expanded $(1-3x)^{-1}$ and obtained:

$$1 - 3 x + 9 x^2 - 27 x^3 + 81 x^4 - 243 x^5 + 729 x^6$$

Then I took the sequence of $x^2$ which is $(0,0,1,0,0,\ldots)$ and then multitplied both sequences and obtained:

$$(0,0,1,-3,9,-27,81,\ldots)$$

And here I don't understand why the answer is $(0,0,1,-3,3^2,-3^3)$. I may have made some silly mistake but I can't spot it.

Answer 1

Hint: $$ \frac1{1-x}=1+x+x^2+x^3+\dots $$