# Ways of calculating Z Transform of a geometric serie

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

• Hint: property of displacement – manooooh May 22 at 17:06
• Welcome to math.SE!! – manooooh May 22 at 17:06
• Thanks ! That hint doesn't help me a lot since I'm not that good in maths and I do not know where to look – Exania May 22 at 17:25