$\mathcal{F}$ is the family of all $n$-tuples $(A_1,A_2,\ldots,A_n)$, where $A_i$ is a subset of a set containing $1998$ elements. We can easily find that $|\mathcal{F}|=2^{1998n}$. $\mathcal{P}$ is the family of all $n$-tuples $(A_1,A_2,\ldots,A_n)$, where $A_i$ is a subset of a set containing $1997$ elements. We also have $|\mathcal{P}|= 2^{1997n}$. For each $x \in X=\{1,2,\ldots,1998\}$, the number of $n$-tuples $(A_1,A_2,\ldots,A_n)$, where $A_i$ is a subset of $X$, and with $x\in A_1 \cup A_2 \cup \cdots \cup A_n$ is $|\mathcal{F}|-|\mathcal{P}|$. So, the sum we have to determine is $1998(|\mathcal{F}|-|\mathcal{P}|)$.

Let $n, N\in \mathbb{N}^*$, and let $\mathcal{F}$ be the family of all $n$-tuples $(A_1, A_2, \ldots, A_n)$ such that each $A_{i}$ is a subset of $\{1, 2, \ldots, N\}$. Let $|A|$ denote the number of elements of the set $A$. Find \[ \sum_{(A_1, A_2, \ldots, A_n)\in \mathcal{F}} |A_1\cup A_2\cup \cdots \cup A_n| .\] There are $\binom {N} {k}$ subsets $A \subseteq \{1,2,\ldots,N\}$ with $|A| = k$, $0\leq k \leq N$. For each such $A$, the number of elements $(A_1, A_2, \ldots, A_n) \in \mathcal{F}$ with $A_1\cup A_2\cup \cdots \cup A_n = A$ is obviously $(2^{n} - 1)^k$. Therefore the sum we seek is \[ \sum_{k=1}^N \binom {N} {k} k(2^{n} - 1)^k = N(2^n-1)\sum_{k=1}^{N} \binom {N-1} {k-1} (2^{n} - 1)^{k-1} = N(2^n-1)2^{n(N-1)} .\] Most likely there must exist a proof by generating functions.

mavropnevma wrote: Most likely there must exist a proof by generating functions. Define $F_{m,n}(t,x):=\left(1+t\left((1+x)^n-1\right)\right)^m$. The coefficient of $t^k x^\ell$ is the number of $n$-tuples $\left(A_1,A_2,\ldots,A_n\right)$ of subsets of $\{1,2,\ldots,m\}$ such that $\bigcup_{i=1}^n A_i$ has $k$ elements and that $\sum_{i=1}^n \left|A_i\right|=\ell$. We want to associate $t^k$ with $k$, and this can be done via differentiating $F_{m,n}(t,1)$ with respect to $t$ and then replacing $t$ by $1$. The sum of $\left|\bigcup_{i=1}^n A_i\right|$ is then \[\left.\frac{\partial}{\partial t} F_{m,n}(t,1) \right|_{t:=1} = m \left(2^n-1\right)2^{(m-1)n}\,.\] This solution is not particularly elegant, but if one is asked to determine the sum of $\left(\sum_{i=1}^n \left|A_i\right|\right) \left|\bigcup_{i=1}^n A_i\right|$, then one can differentiate $F_{m,n}(t,x)$ with respect to each of its variables: \begin{align*}\left.\frac{\partial^2}{\partial x \partial t} F_{m,n}(t,x) \right|_{t:=1,x:=1} &= \left.\frac{\partial}{\partial x} m \left((1+x)^n-1\right)(1+x)^{(m-1)n}\right|_{x=1} \\ &=mn2^{(m-1)n-1}\left(m2^{n}-(m-1)\right)\,.\end{align*} And this is exactly the value of the sum.

shobber wrote: Let $F$ be the set of all $n$-tuples $(A_1, \ldots, A_n)$ such that each $A_{i}$ is a subset of $\{1, 2, \ldots, 1998\}$. Let $|A|$ denote the number of elements of the set $A$. Find \[ \sum_{(A_1, \ldots, A_n)\in F} |A_1\cup A_2\cup \cdots \cup A_n| \] For an element $x \in \{1,2,\cdots,1998\}$, the probability that $x \in |A_1\cup A_2\cup \cdots \cup A_n| $ is clearly $1 - \frac{1}{2^n}$. Thus the expected cardinality of $|A_1\cup A_2\cup \cdots \cup A_n| $ is $1998 \left ( 1 - \frac{1}{2^n} \right ) $. The summation evaluates to $$1998 \left ( 1 - \frac{1}{2^n} \right ) \cdot |F| = 1998 \left ( 1 - \frac{1}{2^n} \right ) \cdot 2^{1998n}$$

For each element, there is a $\frac 12$ chance it is not in $A_i$. Thus the probability that a given element is in at least one set is $1-\frac{1}{2^n}$, so the expected value of \[| A_1 \cup A_2 \cup \ldots \cup A_n | = 1998 \left(1 - \frac{1}{2^n}\right).\] As there are $2^{1998n}$ $n$-tuples in $F$, our total sum is $\boxed{2^{1998n} \cdot 1998 \cdot \left(1 - \frac{1}{2^n}\right)}$.

Let $\mathbb{I}\{E\}$ be the indicator of the proposition $E$, i.e. $\mathbb{I}\{E\}\in\{0,1\}$ and $\mathbb{I}\{E\}=1$ iff $E$ is true. Set $N\triangleq 1998$ and observe that \begin{align*} &\sum_{(A_1,\dots,A_n)\in F} \bigl|A_1\cup A_2\cup \cdots \cup A_n\bigr| \\ & = \sum_{(A_1,\dots,A_n)\in F} \sum_{k=1}^N \mathbb{I}\bigl\{k\in A_1\cup \cdots \cup A_n\bigr\} \\ &=\sum_{k=1}^N \sum_{(A_1,\dots,A_n)\in F} \mathbb{I}\bigl\{k\in A_1\cup \cdots \cup A_n\bigr\} \\ &= N\sum_{(A_1,\dots,A_n)\in F} \mathbb{I}\bigl\{k\in A_1\cup \cdots \cup A_n\bigr\}, \end{align*}from symmetry. Now, there are $(2^N)^n =2^{Nn}$ $n$-tuples $(A_1,\dots,A_n)\in F$. Of all those, $(2^{N-1})^n$ do not contain $k$. So, the number of those $n$-tuples containing $k$ is \[ 2^{Nn}-2^{(N-1)n} = 2^{(N-1)n}(2^n-1), \]yielding the desired answer as $N2^{(N-1)n}(2^n-1)$.

Nice use of the Indicator function !

Note that there are $2^{1998}$ different subsets. Thus, there are $2^{1998n}$ different $n$-tuples of $F$. Consider any element $s$. The probability that $s$ does not appear in $A_i$ is $\tfrac{1}{2}$. The probability is doesn't appear in $F$ is $\tfrac{1}{2^n}$. Therefore, the probability is appears in $F$ is $1 - \tfrac{1}{2^n}$. Hence, the desired sum is equivalent to \begin{align*} 2^{1998n} \cdot 1998 \cdot \left(1 - \tfrac{1}{2^n}\right) = \boxed{1998 \cdot 2^{1998n} - 1998 \cdot 2^{1997n}}. \end{align*}

Consider the element $1$. We take how many times it appears in the summation and then multiply by $1998$ for our solution. It appears iff some set contains the element $1$. By complementary counting, this occurs in ${2^{1998}}^n - {2^{1997}}^n$, so our answer is $\boxed{1998(2^{1998n} - 2^{1997n})}$

For a specific tuple $A = (A_1, \dots, A_n)$, let \[X_{i, A} = \begin{cases}1 & \text{if }i \in A_1 \cup A_2 \cup \dots \cup A_n \\ 0 & \text{otherwise}\end{cases}\]for each $1 \leq i \leq n$. Now \[\sum_{A \in F} |A_1 \cup A_2 \cup \dots \cup A_n| = \sum_{A \in F} \sum_{i = 1}^n X_{i, A} = \sum_{i = 1}^n \sum_{A \in F} X_{i, A}.\]But for a fixed $i$, $(2^{1998})^n - \sum_{A \in F} X_{i, A}$ counts the number of tuples $A$ for which $i \notin A_j$ for any $1 \leq j \leq n$, which is $(2^{1997})^n$. It follows that \[\sum_{A \in F} X_{i, A} = 2^{1998n} - 2^{1997n}\]and thus that our desired sum is $\boxed{1998(2^{1998n} - 2^{1997n})}$.