1. Let $A=\{1,2, \cdots, n\}$, then the sum of the elements in all non-empty proper subsets of $A$ is
1. $\frac{n(n+1)\left(2^{n-1}-1\right)}{2}$ Explanation: The number of times each element appears is the number of subsets of the remaining $n-1$ elements minus 1, i.e., $2^{n-1}$
$$
-1, \therefore S_{n}=\left(2^{n-1}-1\right)(1+2+\cdots+n)=\frac{n(n+1)\left(2^{n-1}-1\right)}{2}
$$