Question 6 Let $A \subset S$ and $A \neq \varnothing$, we call
$$
D_{\mathbf{A}}=\frac{1}{|A|} \sum_{\mathbf{a} \in \mathrm{A}} a_{\mathbf{.}}
$$

the average value of $A$. Find the arithmetic mean of all $D_{\mathrm{A}}$.
$$
\begin{array}{ll}
\text { Let } P(S)=\{A \mid A \subset S\} . \\
& R_{\mathrm{E}}=\{A \subset S|| A \mid=k\}, k=1,2 \\
\cdots, n . \\
\text { Then, } \quad R_{1} \cup R_{2} \cup \cdots \cup R_{\mathrm{n}}=P(S)-\{\varnothing\}, \\
& R_{1} \cap R_{1}=\varnothing, 1 \leqslant i<1 \leqslant n .
\end{array}
$$

Then, $\quad R_{1} \cup R_{2} \cup \cdots \cup R_{\mathrm{n}}=P(S)-\{\varnothing\}$,

So, $\sum_{\substack{\mathrm{A} \subset \mathrm{S} \\ \mathrm{A} \neq \varnothing}} D_{\mathrm{A}}=\sum_{\mathrm{k}=1}^{\mathrm{D}} \sum_{\mathrm{A} \in \mathrm{R}_{\mathrm{B}}} D_{\mathrm{A}}$
$$
=\sum_{k=1}^{\mathrm{n}}\left(\frac{1}{k} \sum_{A \in R_{k}} \sum_{a \in A} a\right) .
$$

Now, take any $i \in S$, because the number of $A$ such that $i \in A \in R_{\mathbf{k}}$ is $C_{n-1}^{k-1}$, the sum $\sum_{A \in R_{k}} \sum_{a \in A} a$ contributes $i \mathrm{C}_{\mathrm{u}-1}^{\mathrm{k}-1}$. Therefore,
$$
\begin{array}{l}
\sum_{A \in R_{k}} D_{A}=\frac{1}{k} \sum_{A \in R_{k}} \sum_{a \in A} a \\
\frac{1}{k}(1+2+\cdots+n) C_{n-1}^{k-1}=\frac{1}{2}(n+1) C_{n}^{k} .
\end{array}
$$

Thus, $\sum_{\substack{\mathrm{A} \subset \mathrm{S} \\ \mathrm{A}+\varnothing}} D_{\mathrm{A}}=\frac{1}{2}(d+1) \sum_{n=1}^{\mathrm{D}} C_{\mathrm{B}}^{\mathrm{k}}$
$$
=\frac{1}{2}(n+1)\left(2^{\mathrm{n}}-1\right) \text {. }
$$

Hence, $\sum_{\substack{A \subset S \\ A \neq \varnothing}} D_{\mathrm{A}} /\left(2^{\mathrm{n}}-1\right)=\frac{1}{2}(n+1)$,
which means the arithmetic mean of the averages of all non-empty subsets of $S$ is $\frac{n+1}{2}$.