If you have a function $x(n) = 2^nu(n+1)$
And $u(n) = 1 \quad if \quad n > 0$
And you need the Z-Transform you'll have to go through a sum, knowing:
$$\sum_{n=0}^\infty 2^nz^{-n} = \frac{1}{1-{2z^{-1}}}$$
But since we have $u(n+1)$ in the function, the sum will start from $-1$
I know that we can solve the sum like that :
$$\sum_{n=-1}^\infty 2^nz^{-n} = \frac{z}{2} + \frac{z}{2}*\frac{2}{z} + \frac{z}{2}*(\frac{2}{z})^2 +...+\frac{z}{2}*(\frac{2}{z})^\infty$$
$$\sum_{n=-1}^\infty 2^nz^{-n} = \frac{z}{2}\sum_{n=0}^\infty 2^nz^{-n} = \frac{z}{2}*\frac{1}{1-2z^{-1}}$$
But couldn't we just get one term out of it :
$$\sum_{n=-1}^\infty 2^nz^{-n} = \frac{z}{2} + \sum_{n=0}^\infty 2^nz^{-n}=\frac{z}{2} +\frac{1}{1-2z^{-1}}$$
Is the last sum not possible? If not which theorem could explain it? Thanks for the help