Summation equation for $2^{x-1}$ Since everyone freaked out, I made the variables are the same.
$$
\sum_{x=1}^{n} 2^{x-1}
$$
I've been trying to find this for a while. I tried the usually geometric equation (Here) but I couldn't get it right (if you need me to post my work I will). Here's the outputs I need:
1, 3, 7, 15, 31, 63
If my math is correct.

In this instance, without explicitly using the formula for geometric series,
$$\begin{align*}
\sum_{x=1}^n 2^{x-1} &= 1 + 2 + 2^2 + 2^3 + \cdots + 2^{n-1}\\
&= 1 + (1 + 2 + 2^2 + 2^3 + \cdots + 2^{n-1}) - 1 &\text{add and subtract}~ 1\\
&= (1 + 1) + (2 + 2^2 + 2^3 + \cdots + 2^{n-1}) - 1 &\text{regroup}\\
&= 2 + (2 + 2^2 + 2^3 + \cdots + 2^{n-1}) - 1\\
&= (2 + 2) + (2^2 + \cdots + 2^{n-1}) - 1 &\text{regroup again}\\
&= 2^2 + (2^2 + 2^3 + \cdots + 2^{n-1}) - 1\\
&= (2^2 + 2^2) + (2^3 + \cdots + 2^{n-1}) - 1 &\text{regroup again}\\
&= 2^3 + (2^3 + \cdots + 2^{n-1}) - 1\\
&= \cdots &\text{lather, rinse, repeat}\\
&= 2^{n-1} + (2^{n-1}) - 1 &\text{nearly done}\\
&= 2^n - 1.
\end{align*}$$
Now that we know the form of the result, it is also possible
to prove the result

$$\sum_{x=1}^n 2^{x-1} = 2^n - 1$$

more formally by induction. Clearly, 
 the result holds when $n = 1$ since $2^0 = 2^1 - 1$.
Then, if the result holds for some positive integer $n$, we have
that
$$\sum_{x=1}^{n+1} 2^{x-1} = \sum_{x=1}^n 2^{x-1} + 2^n
= (2^n-1) + 2^n = 2^{n+1} - 1$$
and so the result holds for $n+1$ as well. Since we know
that the result holds when $n=1$, it follows by induction
that it holds for all positive integers $n$.
