Answer
\[
\boxed{
s_k
= \begin{cases}
2^{\,k/2}, & \text{if \(k\) is even},\\[4pt]
2^{\,\tfrac{k+1}{2}}, & \text{if \(k\) is odd}.
\end{cases}
}
\]
Equivalently,
\[
\boxed{s_k = 2^{\,\lfloor (k+1)/2 \rfloor}, \quad k \ge 0.}
\]
Work Step by Step
We have the sequence \(\{s_k\}\) defined by
\[
\begin{cases}
s_0 = 1, \quad s_1 = 2,\\
s_k = 2\,s_{k-2} \quad \text{for all integers } k \ge 2.
\end{cases}
\]
---
## (a) Use Iteration to Guess an Explicit Formula
Compute the first few terms:
\(s_0 = 1\).
\(s_1 = 2\).
\(s_2 = 2\,s_0 = 2 \times 1 = 2.\)
\(s_3 = 2\,s_1 = 2 \times 2 = 4.\)
\(s_4 = 2\,s_2 = 2 \times 2 = 4.\)
\(s_5 = 2\,s_3 = 2 \times 4 = 8.\)
\(s_6 = 2\,s_4 = 2 \times 4 = 8.\)
\(s_7 = 2\,s_5 = 2 \times 8 = 16.\)
\(s_8 = 2\,s_6 = 2 \times 8 = 16.\)
We see that:
- **Even indices**: \(s_0 = 1,\; s_2 = 2,\; s_4 = 4,\; s_6 = 8,\; s_8 = 16,\dots\)
This suggests \(s_{2n} = 2^n\).
- **Odd indices**: \(s_1 = 2,\; s_3 = 4,\; s_5 = 8,\; s_7 = 16,\dots\)
This suggests \(s_{2n+1} = 2^{\,n+1}.\)
Hence a natural piecewise explicit form is:
\[
\boxed{
s_k
= \begin{cases}
2^{k/2}, & \text{if \(k\) is even},\\[6pt]
2^{(k+1)/2}, & \text{if \(k\) is odd}.
\end{cases}
}
\]
Alternatively, one can unify these two cases using the floor function:
\[
\boxed{
s_k = 2^{\,\lfloor (k+1)/2 \rfloor}, \quad \text{for all integers } k \ge 0.
}
\]
Either version is a correct closed‐form description of \(s_k\).
---
## (b) Verification by Strong Mathematical Induction
We will prove the piecewise version
\[
s_{2n} = 2^n
\quad\text{and}\quad
s_{2n+1} = 2^{\,n+1}
\]
for all \(n \ge 0\). Equivalently, this shows the same result as the floor‐function formula.
### Base Cases
- **\(k = 0\)** (even): The formula says \(s_0 = 2^{\,0} = 1\). This matches \(s_0 = 1\).
- **\(k = 1\)** (odd): The formula says \(s_1 = 2^{\,0+1} = 2\). This matches \(s_1 = 2\).
### Inductive Step
Assume that for **all** integers \(j < k\), the formulas hold. We must show it holds for \(k\).
1. **Case 1: \(k\) is even, say \(k = 2n\).**
Then \(k - 2 = 2(n-1)\). By the inductive hypothesis,
\[
s_{k-2}
= s_{2(n-1)}
= 2^{\,n-1}.
\]
The recursion says \(s_k = 2\,s_{k-2}\). Thus
\[
s_{2n}
= 2 \, s_{2(n-1)}
= 2 \times 2^{\,n-1}
= 2^n,
\]
which matches the even‐index formula.
2. **Case 2: \(k\) is odd, say \(k = 2n+1\).**
Then \(k - 2 = 2n - 1 = 2(n-1) + 1\). By the inductive hypothesis,
\[
s_{k-2}
= s_{\,2(n-1) + 1}
= 2^{\, (n-1) + 1}
= 2^n.
\]
Again using \(s_k = 2\,s_{k-2}\),
\[
s_{2n+1}
= 2 \, s_{\,2(n-1) + 1}
= 2 \times 2^n
= 2^{\,n+1},
\]
matching the odd‐index formula.
By strong induction, these formulas hold for all \(k \ge 0\).
