Question 5 Let the set $S=\{1,2, \cdots, 1000\}$. Now for any non-empty subset $A$ of $S$, let $\alpha_{\mathrm{A}}$ denote the sum of the largest and smallest numbers in $A$, then, the arithmetic mean of all such $\alpha_{\Delta}$ is $\qquad$
The answer is 1001.
Let $S=\{1,2, \cdots, n\}, n \in N$. Let $m_{A}, M_{A}$ represent the minimum and maximum elements of a non-empty subset $A$ of $S$, respectively. Then the arithmetic mean of all $\alpha_{A}$ is
$$
\frac{1}{2^{n}-1} \sum_{\substack{A \subset S \\ A \neq \varnothing}}\left(m_{A}+M_{A}\right).
$$

Since for $1 \leqslant k \leqslant n$, there are $2^{n-k}$ subsets $A$ satisfying $m_{A}=k$, we have
$$
\begin{array}{l}
\sum_{\substack{A \subset S \\ A \neq \infty}} m_{A}=1 \cdot 2^{n-1}+2 \cdot 2^{n-2}+\cdots+ \\
k \cdot 2^{n-1}+\cdots+n \cdot 2^{0}.
\end{array}
$$

And for $1 \leqslant k \leqslant n$, there are $2^{k-1}$ subsets $A$ such that $M_{A}=k$, hence
$$
\begin{array}{l}
\sum_{\substack{A}} M_{A}=n \cdot 2^{n-1}+(n-1) 2^{n-2}+\cdots \\
\quad+(n-k+1) 2^{n-k}+\cdots+1 \cdot 2^{0}.(3)
\end{array}
$$
Adding (2) and (3) gives
$$
\begin{array}{l}
\sum_{A \subseteq S}\left(m_{A}+M_{A}\right)=(n+1)\left(2^{n-1}+2^{n-2}\right. \\
A \neq \varnothing \\
+\cdots+2+1)=(n+1)\left(2^{n}-1\right).
\end{array}
$$

Combining (1), we know that the arithmetic mean of all such $\alpha_{A}$ is $n+1$.
From (2), we get
$$
\begin{array}{l}
2 \sum_{A \subset S} m_{A}=2^{n}+2 \cdot 2^{n-1}+\cdots \\
A \neq \varnothing \\
\quad+(t+1) 2^{n-I}+\cdots+n \cdot 2.
\end{array}
$$
Subtracting (2) from (A), we get
$$
\begin{aligned}
\sum_{\substack{A \subset S}}^{n_{A} \neq \varnothing} & =2^{n}+2^{n-1}+\cdots+2^{n-x}+2-n \\
& =2^{u+1}-n-2.
\end{aligned}
$$

Similarly, we have
$$
\sum_{\substack{A \subset S \\ A \neq \varnothing}} M_{\triangle}=(n-1) 2^{\mathbb{n}}+1
$$

Finally, let $1 \leqslant i<i \leqslant n$, then the number of subsets $A \subset S$ with $i$ and $1$ as the minimum and maximum elements, respectively, is $2^{i-1-1}$, and in this case, $\alpha_{A}=i+j$. Noting that $\alpha_{\{1\}}=2 i, 1 \leqslant i \leqslant n$. Therefore, we have the identity
$$
\begin{array}{l}
\mathbf{2} \sum_{\mathbf{1}=1}^{\mathbf{n}} i+\sum_{1 \leqslant \mathrm{i}<\mathbf{j} \leqslant \mathbf{n}}(i+j) 2^{i-1-1} \\
=(n+1) \cdot\left(2^{\mathbf{n}}-1\right),
\end{array}
$$

or $\sum_{1 \leqslant \mathrm{i}<\mathrm{j}=\mathrm{n}}(i+1) 2^{\mathrm{j}-1}=2(n+1)\left(2^{\mathrm{n}}-n-1\right)$ (8)