Answer
1. **(a)** By iterating and noticing the exponents follow the Fibonacci pattern, we guess
\[
\boxed{u_k \;=\; 2^{\,F_{k+1}}},
\] where \(\{F_n\}\) is the Fibonacci sequence with \(F_1=1,\;F_2=1,\;F_3=2,\dots\).
2. **(b)** A straightforward strong induction confirms that this closed‐form matches both the initial conditions and the multiplicative recurrence.
Work Step by Step
We have the sequence \(\{u_k\}\) defined by
\[
\begin{cases}
u_0 = 2,\quad u_1 = 2,\\[4pt]
u_k \;=\; u_{k-2}\,\cdot\,u_{k-1},\quad \text{for all integers } k \ge 2.
\end{cases}
\]
---
## (a) Use Iteration to Guess an Explicit Formula
Let’s compute the first few terms:
\(u_0 = 2.\)
\(u_1 = 2.\)
\(u_2 = u_0 \cdot u_1 = 2 \times 2 = 4.\)
\(u_3 = u_1 \cdot u_2 = 2 \times 4 = 8.\)
\(u_4 = u_2 \cdot u_3 = 4 \times 8 = 32.\)
\(u_5 = u_3 \cdot u_4 = 8 \times 32 = 256.\)
\(u_6 = u_4 \cdot u_5 = 32 \times 256 = 8192.\)
Now observe that these are all powers of 2:
\[
\begin{aligned}
u_0 &= 2 = 2^1,\\
u_1 &= 2 = 2^1,\\
u_2 &= 4 = 2^2,\\
u_3 &= 8 = 2^3,\\
u_4 &= 32 = 2^5,\\
u_5 &= 256 = 2^8,\\
u_6 &= 8192 = 2^{13},\\
&\;\;\dots
\end{aligned}
\]
Look at the exponents:
\[
1,\;1,\;2,\;3,\;5,\;8,\;13,\;\dots
\]
Those are the Fibonacci numbers, starting with \(F_1=1,\; F_2=1,\; F_3=2,\; F_4=3,\; F_5=5,\;\dots\)
Hence it appears
\[
u_k = 2^{\,F_{k+1}},
\]
where \(F_{n}\) denotes the \(n\)-th Fibonacci number under the standard convention \(F_1=1,\;F_2=1,\;F_3=2,\dots\).
**Thus, our guessed closed‐form is**
\[
\boxed{u_k \;=\; 2^{\,F_{k+1}}.}
\]
---
## (b) Verification by Strong Induction
We will prove that \(u_k = 2^{\,F_{k+1}}\) satisfies both the initial conditions and the recurrence \(u_k = u_{k-2}\,u_{k-1}\).
### Base Cases
- **\(k=0\):** The right‐hand side is \(2^{F_{0+1}} = 2^{F_1} = 2^1 = 2\).
Matches \(u_0=2.\)
- **\(k=1\):** Then \(u_1 \stackrel{?}{=} 2^{\,F_2} = 2^1 = 2.\)
Matches \(u_1=2.\)
Hence the formula holds for \(k=0\) and \(k=1\).
### Inductive Step
Assume for all integers \(j
