What is the sum of series in terms of z and N of this expression?

Question

$$\sum_{n=1}^{N}\frac {z^n}{2^n}$$

I attempted to use the method of difference but it didn't work.

Answer 1

Note that you can write $$1+x+x^2+...+x^N=(1+x+x^2+...+x^N)(1-x)/(1-x)=(1-x^{N+1})/(1-x)$$ if $x\ne 1$

Answer 2

Use the formula for the sum of consecutive terms of a geometric progression: $$\frac z2\,\frac{\Bigl(\dfrac z2\Bigr)^{\!N}-1}{\dfrac z2-1}=\frac{z\Bigl(z^N-2^N\Bigr)}{2^N(z-2)}\qquad(z\ne 2)$$