3 Let the set $S=\{1,2,3, \cdots, n\}$, for each non-empty subset $S_{i}$ of $S$, define its "weight" as the sum of all elements in $S_{i}$, then the arithmetic mean of the "weights" of all non-empty subsets of $S$ is $\qquad$ .
$3 \frac{n(n+1) \cdot 2^{n-2}}{2^{n}-1}$. Solution 1 Define the "weight" of the empty set $\varnothing$ as 0. The sum of the "weight" of each subset $S_{i}$ and its complement $S_{i}$ is $1+2+3+\cdots+n=\frac{n(n+1)}{2}$, therefore the total "weight" of all non-empty subsets $=\frac{n(n+1)}{2} \cdot \frac{2^{n}}{2}=n(n+1) \cdot 2^{n-2}$ (since the total number of subsets of $S$ is $2^{n}$), so the arithmetic mean of the "weights" $=\frac{n(n+1) \cdot 2^{n-2}}{2^{n}-1}$.
Solution 2 Each element appears $2^{n-1}$ times, which is the number of all subsets (including the empty set $\varnothing$) of a "set with $n-1$ elements", therefore the arithmetic mean of the "weights" $=\frac{(1+2+3+\cdots+n) \cdot 2^{n-1}}{2^{n}-1}=$ $\frac{n(n+1) \cdot 2^{n-2}}{2^{n}-1}$