finding inverse Z transform with usage of tables (LTI and causal sequences) problem is as follows

try to find the inverse Z 

$Z^{-1} (\frac{3}{z+2}) = ???$

with the usage of z-transform tables



Ok, so in order to find something from the table, I thought that we expand with z and try to see if we  get closer to any kind of expression in the tables

$Z^{-1} (\frac{3z}{z*(z+2)}) = Z^{-1} (3*\frac{1}{z}\frac{z}{z+2}) $

then we get constant out

$=3*Z^{-1}(z^{-1}*\frac{z}{z+2})$

then we see that right side fraction can be found in the table, but the combined sum expression can also be found with further manipulation (as can be seen later)

So, I think we declare temp variable here to clean it up (this portion is found in the table green portion)

$F(z) = \frac{z}{z-(-2)} \iff f[k]=(-2)^k $

then we can make it so

$3*Z^{-1}(z^{-1}*F(z))$

then we find the direct expression in the table (we have delay of one unit because of $z^{-1}$ term, it is the orange portion)

$3*Z^{-1}(z^{-1}*F(z)) = 3* u[n-1]*f[n-1]$

so it looks like answer should be

$3* u[n-1]*f[n-1] = 3*u[n-1]*(-2)^{n-1}$

my teacher hasnt provided the solutions guide yet, despite that he promised to do so, and I reminded him about it, so I was just checking if this is the right solution and thinking... 

I did the same problem, but I used polynomial long-division style to get the following series representation

$3*z^{-1}-6*z^{-2}+12*z^{-3}-24*z^{-4}...$

so based on that I guess that the $x[n]=3*(-1)^n*2^n $

where $ n&gt;=0$

so, I was just a little bit confused does there actually exist the delay of one unit in the original input sequence (result of inverse Z), sometimes I get confused about the causal sequence when you multiply with the unit step sequence, but then, it seems that sometimes also it can just be stated that the original sequence must have been causal sequence, so you just define n>=0 or something like that???
Your solution is correct. This is the way I see it (assuming that the sequence is causal/right-sided):

$$\mathcal{Z}^{-1}\left\{\frac{3}{z+2}\right\}[n]=3\mathcal{Z}^{-1}\left\{z^{-1}\frac{z}{z+2}\right\}[n]=\\=3\mathcal{Z}^{-1}\left\{\frac{z}{z+2}\right\}[n-1]=3(-2)^{n-1}u[n-1]$$