continuous signals - z-transform of $2^k$

It seems that you can decompose it as such:

$f(n) = a^n u(n) + a^{-n} u(-n-1)$

But I already have issue here,

is it basically saying that $ u(n) + u(-n-1) = 1$?

this is the plot of u(n) and u(-n-1):

if you sum them, wouldn't you just basically combine them? So you get 1 from $- \infty$ to -1, and from 0 to $\infty$

between n = -1 and 0 it's 0

no?

I know that the answer to z-transform $2^k$ is $\frac{z}{z-2}$ but how do you arrive at this?

general formula of z-transform is:

$F(z) = \sum_{k=\infty}^{0} f[k] z^{-k} $

applying formula:

$F(z) = \sum_{k=\infty}^{0} 2^k z^{-k} $

it would look like this, no?