<s>[INST] From $32\cdot (\frac{1-.5^n}{1-.5})$ to $\frac{64(2^n-1)}{2^n}$ I have to prove that  $32\cdot (\frac{1-.5^n}{1-.5})$ is equal to $\frac{64(2^n-1)}{2^n}$
I know that they are equal because when I put them in my calculator the graphs are the same. However, I'd like to do this manually.
So far I have done:
$$32\cdot (\frac{1-.5^n}{1-.5}) = 32 \cdot (\frac{.5}{.5} + \frac{.5^n}{.5}) = 32 \cdot (\frac{.5\cdot.5^n}{.5}) = \frac{32\cdot.5\cdot.5^n}{.5}= \frac{8^n}{.5} $$
How do I solve this?
 [/INST] Easier to start from the right. 
$$\begin{align}\frac {64(2^n-1)}{2^n}
&=64\left(\frac {2^n}{2^n}-\frac 1{2^n}\right)\\\\
&=\frac {32}{0.5}\left(1-\left(\frac 12\right)^n\right)\\\\
&=\frac{32(1-0.5^n)}{1-0.5}
\end{align}$$
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