$1 \cdot 5 n$ is a fixed positive integer, find the sum of all positive integers that have the following property: in binary, this number has exactly $2 n$ digits, of which $n$ are 1s and $n$ are 0s (the leading digit cannot be $0$).
[Solution] The first digit of such a positive integer must be 1, and among the other $2n-1$ digits, exactly $n$ digits are 0, so there are $C_{2 n-1}^{n}$ such numbers. The 1 in the $k$-th position from the right $(1 \leqslant k \leqslant 2 n-1)$ represents $2^{k-1}$, and appears $C_{2 n-2}^{n}$ times (except for the $k$-th position from the right and the $2n$-th position, the remaining $2 n-2$ positions have exactly $n$ digits that are 0), so the total sum of such positive integers is
$$
\begin{aligned}
& 2^{2 n-1} \times C_{2 n-1}^{n}+\left(2^{2 n-2}+\cdots+2+1\right) \times C_{2 n-2}^{n} \\
= & 2^{2 n-1} \times C_{2 n-1}^{n}+\left(2^{2 n-1}-1\right) \times C_{2 n-2}^{n} .
\end{aligned}
$$