Answer
See explanation
Work Step by Step
We have the recurrence
\[
\begin{cases}
a_0 = 2,\\
a_k \;=\; \dfrac{a_{k-1}}{2\,a_{k-1} \;-\; 1}\quad\text{for all integers }k \ge 1.
\end{cases}
\]
Let us first compute the first few terms to look for a pattern.
1. **\(k = 0\)**:
\(a_0 = 2.\)
2. **\(k = 1\)**:
\[
a_1 \;=\; \frac{a_0}{2\,a_0 - 1}
= \frac{2}{2\cdot 2 \;-\; 1}
= \frac{2}{4 - 1}
= \frac{2}{3}.
\]
3. **\(k = 2\)**:
\[
a_2 \;=\; \frac{a_1}{2\,a_1 - 1}
= \frac{\tfrac{2}{3}}{2\cdot \tfrac{2}{3} \;-\; 1}
= \frac{\tfrac{2}{3}}{\tfrac{4}{3} - 1}
= \frac{\tfrac{2}{3}}{\tfrac{1}{3}}
= 2.
\]
4. **\(k = 3\)**:
\[
a_3
= \frac{a_2}{2\,a_2 - 1}
= \frac{2}{2\cdot 2 - 1}
= \frac{2}{3}.
\]
We see the terms so far are:
\[
a_0 = 2,\quad a_1 = \tfrac{2}{3},\quad a_2 = 2,\quad a_3 = \tfrac{2}{3},\dots
\]
It repeats with period 2:
- Even \(k\): \(a_k = 2.\)
- Odd \(k\): \(a_k = \frac{2}{3}.\)
---
## (a) Iteration and a “Guess” for the Explicit Formula
From the first few terms, we can **guess** that the sequence alternates between 2 and \(\tfrac{2}{3}\).
Hence, the simplest way to write the closed‐form is a piecewise function:
\[
\boxed{
a_k
= \begin{cases}
2, & \text{if \(k\) is even},\\[6pt]
\tfrac{2}{3}, & \text{if \(k\) is odd}.
\end{cases}
}
\]
That is our “explicit formula” for all integers \(k \ge 0\).
---
## (b) Verification by Strong (or Simple) Induction
We can prove this formula is correct for all \(k\) by induction on \(k\).
### Base Cases
- **\(k=0\)** (even):
The formula says \(a_0 = 2\).
Indeed, the problem statement gives \(a_0=2\).
- **\(k=1\)** (odd):
The formula says \(a_1 = \tfrac{2}{3}\).
We already computed \(a_1 = \tfrac{2}{3}\) directly from the recurrence.
So the base cases hold.
### Inductive Step
Assume for all integers \(n < k\), the formula holds. We want to show it holds for \(k\).
- If **\(k\) is even**, then \(k-1\) is odd. By the inductive hypothesis, \(a_{k-1} = \tfrac{2}{3}\). Then
\[
a_k
\;=\; \frac{a_{k-1}}{2\,a_{k-1} \;-\; 1}
\;=\; \frac{\tfrac{2}{3}}{2\cdot\tfrac{2}{3} \;-\; 1}
\;=\; \frac{\tfrac{2}{3}}{\tfrac{4}{3} - 1}
\;=\; \frac{\tfrac{2}{3}}{\tfrac{1}{3}}
\;=\; 2.
\]
This matches the piecewise formula for \(k\) even.
- If **\(k\) is odd**, then \(k-1\) is even. By the inductive hypothesis, \(a_{k-1} = 2\). Then
\[
a_k
\;=\; \frac{a_{k-1}}{2\,a_{k-1} \;-\; 1}
\;=\; \frac{2}{2\cdot 2 - 1}
\;=\; \frac{2}{3},
\]
which matches the piecewise formula for \(k\) odd.
Thus, in either case, the formula is satisfied for \(k\). By induction, the piecewise formula holds for all \(k \ge 0\).
