The question is to find the z-transform of $$ x(n) = 3n\cdot u(n-2)$$
So far I have seen a question asking to find z-transformation of $x(n)=3^n u(n-2)$ and I know the solution. Howver for the sequence $ x(n) = 3n\cdot u(n-2)$ , I am having a hardtime finding the z-transform. Is this a typo? If not, please provide a solution.
Let us do it step-by-step:
z-transform of $u(n)$: $$u(n)\to \frac{1}{1-z^{-1}}$$
Then for $u(n-2)$ we use time-shifting property:
$$u(n-2)\to z^{-2}\,\frac{1}{1-z^{-1}}=\frac{z^{-2}}{1-z^{-1}}$$
Now it's time to multiply $n$ and use differentiation in z-domain property:
$$nu(n-2)\to -z\frac{d}{dz}\left(\frac{z^{-2}}{1-z^{-1}}\right)$$
And of course there is a $3$ to be multiplied:
$$3nu(n-2)\to -3z\frac{d}{dz}\left(\frac{z^{-2}}{1-z^{-1}}\right)$$
and by simplifying the right-hand-side we get:
$$\mathcal{Z}(3nu(n-2))=3\,\frac{2z^{-2}-z^{-3}}{(1-z^{-1})^2}$$