20.3 .13 Let $F$ be the set of all ordered $n$-tuples $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$, where $A_{i}(1 \leqslant i \leqslant n)$ are subsets of $\{1,2, \cdots, 1998\}$. Find $\sum_{\left(A_{1}, A_{2}, \cdots, A_{n}\right) \in F}\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|$.
For any $x \in\{1,2, \cdots, 1998\}$, let $F_{x}$ be the set of all ordered $n$-tuples $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$ such that $x \notin \bigcup_{i=1}^{n} A_{i}$. Then $|F|=2^{1998 n},\left|F_{x}\right|=2^{1997 n}$, so there are $2^{1998 n}-$ $2^{1997 n}$ $n$-tuples $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$ such that $x \in \bigcup_{i=1}^{n} A_{i}$. Therefore,
$$
\text { the original expression }=\sum_{x \in\{1,2, \cdots, 1998\}}\left(2^{1998 n}-2^{1997 n}\right)=1998 \cdot\left(2^{1998 n}-2^{1997 n}\right) .
$$

Comment: This problem essentially considers how many times a fixed $x \in\{1,2, \cdots, 1998\}$ appears in the summation. Each appearance is recorded as a “ +1 ”, and non-appearance is recorded as a “ 0 ”. The final expression is the count of “ +1 ”. This method is called the “contribution method”.