We have $g(k)=\{ [(1/5)^k]u(k)\text{ for $1 \le k\le3$ and $0$ for other }k\}$
The input is $x(k)=\delta(k) +3\delta(k-1)+ \delta(k-2) $
Using Z transform we have to find the output $y(k)$ and the region of convergence of $Y(z)$.
The problem is that although I have read about the input and output and impulse in the Alan V. Oppenheim book about signals,I dont know how to solve this one,especially the part about finding the output.
In this case using the $\mathcal{Z}$-transform seems almost like an overkill, but let's see. I use $a=1/5$:
$$G(z)=\sum_{k=0}^3a^kz^{-k}=\frac{1-(az^{-1})^4}{1-az^{-1}}\\ X(z)=1+3z^{-1}+z^{-2}$$
So we have
$$Y(z)=X(z)G(z)=G(z)+3G(z)z^{-1}+G(z)z^{-2}\tag{1}$$
Since $Y(z)$ has the same pole as $G(z)$ (which is $z_{\infty}=a$), its region of convergence is $|z|>|a|=1/5$. (Since $Y(z)$ corresponds to a causal sequence, the region of convergence is the region outside a circle).
The inverse transform of (1) is simply
$$y(k)=g(k)+3g(k-1)+g(k-2)$$
I'm sure you can fill in the numerical values yourself if necessary.