Answer
\[
\boxed{
b_k
= \begin{cases}
1, & \text{if \(k\) is odd},\\[4pt]
2, & \text{if \(k\) is even}.
\end{cases}
}
\]
Work Step by Step
We have the sequence \(\{b_k\}\) defined by
\[
\begin{cases}
b_1 = 1,\\[6pt]
b_k = \dfrac{2}{b_{k-1}}, \quad \text{for all integers } k \ge 2.
\end{cases}
\]
---
## (a) Use iteration to guess an explicit formula
Let’s compute the first few terms:
1. \(b_1 = 1.\)
2. \(b_2 = \dfrac{2}{b_1} = \dfrac{2}{1} = 2.\)
3. \(b_3 = \dfrac{2}{b_2} = \dfrac{2}{2} = 1.\)
4. \(b_4 = \dfrac{2}{b_3} = \dfrac{2}{1} = 2.\)
5. \(b_5 = \dfrac{2}{b_4} = \dfrac{2}{2} = 1.\)
Clearly, the terms alternate between \(1\) and \(2\). Thus, by observation:
- If \(k\) is **odd**, \(b_k = 1\).
- If \(k\) is **even**, \(b_k = 2\).
Hence an explicit formula is:
\[
\boxed{
b_k
= \begin{cases}
1, & \text{if \(k\) is odd},\\[6pt]
2, & \text{if \(k\) is even}.
\end{cases}
}
\]
---
## (b) Verification by Strong (or Simple) Induction
We prove that this piecewise formula satisfies the given recurrence.
### Base Case(s)
- For \(k = 1\) (odd), our formula says \(b_1 = 1\). This matches the given initial condition \(b_1 = 1\).
- For \(k = 2\) (even), our formula says \(b_2 = 2\). From the recurrence, \(b_2 = \tfrac{2}{b_1} = \tfrac{2}{1} = 2\). So it matches as well.
### Inductive Step
Assume that for **all** integers \(j\) with \(1 \le j < k\), the formula holds. We want to show it holds for \(k\).
1. If \(k\) is **odd**, then \(k-1\) is even. By the inductive hypothesis, \(b_{k-1} = 2\). The recurrence gives
\[
b_k
= \frac{2}{b_{k-1}}
= \frac{2}{2}
= 1,
\]
which matches our piecewise formula for odd \(k\).
2. If \(k\) is **even**, then \(k-1\) is odd. By the inductive hypothesis, \(b_{k-1} = 1\). The recurrence gives
\[
b_k
= \frac{2}{b_{k-1}}
= \frac{2}{1}
= 2,
\]
which matches our piecewise formula for even \(k\).
Thus, in either case, \(b_k\) agrees with our piecewise definition. By the principle of (strong) mathematical induction, the formula holds for all \(k \ge 1\).
