We can write $\mu = E\sum 2^{r(X_j)-X_j}$, where $X_j$ is the $j$th number drawn and $r(X_j)$ is its rank after ordering. By linearity of the expectation, since the $(X_j,r(X_j))$ have the same distribution, $\mu = k E2^{r(X_1)-X_1}$. To evaluate this expectation, I condition on $X_1$; notice that if $X_1=x$, then I need exactly $r-1$ of the other points $X_2,\ldots ,X_k$ in $[1,x]$ to have $r(X_1)=r$. Thus
$$
\mu= \frac{k}{n-1} \sum_{r=1}^k 2^r \binom{k-1}{r-1} \int_1^n \left(\frac{x-1}{n-1} \right)^{r-1} \left( 1- \frac{x-1}{n-1} \right)^{k-r} 2^{-x}\, dx .
$$
I substitute $t=x-1$, and the sum can be evaluated with the binomial theorem. This gives
$$
\mu = \frac{k}{n-1} \int_0^{n-1} \left( 1 + \frac{t}{n-1}\right)^{k-1} 2^{-t}\, dt .
$$