Fix positive integers $k$ and $n$.Derive a simple expression involving Fibonacci numbers for the number of sequences $(T_1,T_2,\ldots,T_k)$ of subsets $T_i$ of $[n]$ such that $T_1\subseteq T_2\supseteq T_3\subseteq T_4\supseteq\ldots$.

[color=#008000]Moderator says: and the original source for this one is Richard Stanley, [i]Enumerative Combinatorics[/i] vol. 1 (1st edition), exercise 1.15.[/color]
1. **Understanding the Problem:**
   We need to find the number of sequences \((T_1, T_2, \ldots, T_k)\) of subsets \(T_i\) of \([n]\) such that \(T_1 \subseteq T_2 \supseteq T_3 \subseteq T_4 \supseteq \ldots\). Here, \([n]\) denotes the set \(\{1, 2, \ldots, n\}\).

2. **Base Case \(n = 1\):**
   For \(n = 1\), the set \([1] = \{1\}\) has only two subsets: \(\emptyset\) and \(\{1\}\). We define:
   - \(U_k\) as the number of sequences of length \(k\) ending in \(\{1\}\).
   - \(L_k\) as the number of sequences of length \(k\) ending in \(\emptyset\).

3. **Recurrence Relations:**
   - If a sequence \(S\) of length \(2j-1\) ends in \(\emptyset\), it can be extended in three ways: \(\{1\}, \{1\}\), \(\{1\}, \emptyset\), or \(\emptyset, \emptyset\).
   - If \(S\) ends in \(\{1\}\), it can be extended in two ways: \(\{1\}, \{1\}\) or \(\{1\}, \emptyset\).

   From these observations, we derive the following recurrence relations:
   \[
   U_{2j} = U_{2j-1} + L_{2j-1}
   \]
   \[
   L_{2j} = L_{2j-1}
   \]
   \[
   U_{2j+1} = U_{2j-1} + L_{2j-1}
   \]
   \[
   L_{2j+1} = U_{2j-1} + 2L_{2j-1} = U_{2j+1} + L_{2j-1}
   \]

4. **Fibonacci Sequence:**
   We observe that the sequence \(U_1, L_1, U_3, L_3, U_5, L_5, \ldots\) follows the Fibonacci sequence. Specifically:
   \[
   U_1 = 1 = F_1
   \]
   \[
   L_1 = 1 = F_2
   \]
   Thus, the sequence is:
   \[
   U_{2j+1} = F_{2j+1}
   \]
   \[
   L_{2j+1} = F_{2j+2}
   \]

5. **Total Number of Sequences:**
   - For sequences of length \(2j+1\):
     \[
     U_{2j+1} + L_{2j+1} = F_{2j+1} + F_{2j+2} = F_{2j+3}
     \]
   - For sequences of length \(2j\):
     \[
     U_{2j} = U_{2j-1} + L_{2j-1} = F_{2j-1} + F_{2j} = F_{2j+1}
     \]
     \[
     L_{2j} = L_{2j-1} = F_{2j}
     \]
     \[
     U_{2j} + L_{2j} = F_{2j+1} + F_{2j} = F_{2j+2}
     \]

6. **General Case for \(n > 1\):**
   For \(n > 1\), each of the \(n\) elements in \([n]\) can be assigned independently to the sequences \((T_1, T_2, \ldots, T_k)\). Therefore, the total number of sequences is:
   \[
   F_{k+2}^n
   \]

The final answer is \(\boxed{F_{k+2}^n}\).