Confusion applying the shifting property of z-transform I'd like to calculate the z-transform of $Z=(-3)^{-n}u[n-2]=\left(-\dfrac{1}{3}\right)^nu[n-2]$. Also, I'd like to use shifting property, which states:
$$x(n-k)\rightarrow z^{-k}\left(X(z)+\sum_{n=1}^kx(-n)z^k\right)$$
For the function in question, that would be:
$$\left(-\dfrac{1}{3}\right)^nu[n-2] \rightarrow z^{-2}\left(\dfrac{1}{1+\frac{z^{-1}}{3}}-3z+9z^2\right)$$
After some simplification $Z=\dfrac{3}{3z^2+z}-\dfrac{3}{z}+9$. After further simplification verified here, it becomes $Z=\dfrac{27z}{3z+1}$. However, wolfram alpha gives a different transform, which is $\dfrac{1}{9z^2+3z}$. So, I am confused. How should one apply the shifting property? what if the function were $\left(-\dfrac{1}{3}\right)^{n-2}u[n-2]$, or more generally $a^{n-k}u[n-j]$.
 A: Note that there are two versions of $\mathcal{Z}$-transform. The bilateral $\mathcal{Z}$-transform is defined as $$X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n}$$And the unilateral $\mathcal{Z}$-transform is defined as $$X(z) = \sum_{n=0}^{+\infty}x[n]z^{-n}$$Adopting the bilateral transform, yields $$\mathcal{Z}\{\alpha^nu[n]\} = \frac{1}{1-\alpha z^{-1}} \\ \mathcal{Z}\{x[n-n_0]\} = z^{-n_0}X(z)$$
Combining these facts, we have $$\mathcal{Z}\{(-\dfrac{1}{3})^{n-2}u[n-2] \} =  \frac{z^{-2}}{1+\frac{z^{-1}}{3} } \implies \\ \mathcal{Z}\{(-\dfrac{1}{3})^{n}u[n-2] \} = \frac{z^{-2}}{9+3z^{-1} } = \frac{1}{9z^2 +3z} = \frac{1}{3z(3z+1)}$$ This can be easily verified by WA. Also read this documentation which mentions WA computes the unilateral transform by default. The unilateral transform gives $$\mathcal{Z}\{\alpha^nu[n]\} = \frac{1}{1-\alpha z^{-1}} \tag{1}$$ $$\mathcal{Z}\{x[n-1]\} = z^{-1}X(z) + x[-1] \implies \mathcal{Z}\{x[n-2]\} = x[-2] + x[-1]z^{-1} + z^{-2}X(z) \tag{2}$$
By choosing $\alpha=\frac{-1}{3}$ in equation $(1)$, we have $$\mathcal{Z}\{(-\frac{1}{3})^nu[n]\} = \frac{1}{1+\frac{z^{-1}}{3}}$$Now let $x[n] =  (-\frac{1}{3})^nu[n]$, so we have $x[-1] = x[-2] = 0$ which implies, by equation $(2)$, $$\mathcal{Z}\{(-\frac{1}{3})^{n-2}u[n-2]\} = z^{-2}\frac{1}{1+\frac{z^{-1}}{3}}$$
Finally, by linearity property we have $$\mathcal{Z}\{(-\frac{1}{3})^{n}u[n-2]\} = \frac{z^{-2}}{9}\frac{1}{1+\frac{z^{-1}}{3}}$$
In this case both transforms give the same result since $(-\dfrac{1}{3})^{n}u[n] = 0$ for $n\lt0$.
