I am trying to find the origin of $\frac{6}{z-2}$ under the $z$ transformation, i.e. find some $f(k)$ s.t. $f(k)\overset{z}{\rightarrow}F(z)=\frac{6}{z-2}$
According to a formula I have the origin for $\frac{A}{(z-a)^{m+1}}$ is $A\frac{k-m}{k}\binom{k}{m}a^{k-m-1}$ when $k\geq m+1$ and $0$ otherwise.
This gave me the solution $6\cdot2^{k-1}$when $k\geq1$ and $0$ otherwise, but the solution say the answer is $6\cdot2^{k}$ (when $k\geq1$ and $0$ otherwise).
Do I have a mistake ? if so, what is the source of the mistake (e.g. wrong formula, correct formula so I must have a calculation error of some sort etc')