An urn contains n balls numbered $2^{(n-1)}$. Expected value of sum of random selections Here's the problem:
An urn contains n balls numbered $1, 2, 4, 8$, etc. up to $2^{(n-1)}$ where $n > 1$. If a person selects 
1 ball from the urn, then replaces it and selects a second ball, what is the expected value of the sum of the ball's numbers?
So, my line of thinking is this:
Expected value is the sum of all the possible values multiplied by their probabilities. So in this case, the probability of drawing any one ball from the urn is $(1/n)$. The value of the ball is the number written on it, which is $2^{(i-1)}$, if $i$ were the index position in the set of balls.
My thinking is that since there IS replacement, I just need to turn this into a summation, and multiply it by two to get the expected sum of two random draws:
$\begin{equation}
2 * ( \sum_{i = 1}^{n} 2^{i-1}*(1/n) )
\end{equation}$
I'm pretty new to probability stuff, and expected value in particular, so I don't know if this is correct or even if it's close, and would hugely appreciate feedback and/or guidance. Thank you very much.

You are right.  The $\frac 1n$ can be pulled out of the sum, leaving you with a geometric series to sum.
