Answer
See explanation
Work Step by Step
We want to prove by mathematical induction that for all integers \(n \ge 0\), the sequence defined by
\[
\begin{cases}
a_0 = a, \\
a_k = r \, a_{k-1} \quad \text{for all integers } k \ge 1,
\end{cases}
\]
has the closed‐form expression
\[
a_n = a \, r^n.
\]
---
## 1. Base Case \((n = 0)\)
Check the formula for \(n = 0\):
\[
a_0 \stackrel{?}{=} a \, r^0.
\]
Since \(r^0 = 1\), we have
\[
a_0 = a \quad \text{and} \quad a \, r^0 = a \cdot 1 = a,
\]
so \(a_0 = a \, r^0\). Thus, the base case holds.
---
## 2. Inductive Step
Assume that for some integer \(n \ge 0\), the formula holds:
\[
a_n = a \, r^n.
\]
We want to show it also holds for \(n+1\). By the recursive definition,
\[
a_{n+1} = r \, a_n.
\]
Using the inductive hypothesis \(a_n = a \, r^n\), substitute:
\[
a_{n+1} = r \cdot \bigl(a \, r^n\bigr) = a \, r^{n+1}.
\]
Thus, if the formula is true for \(n\), it is also true for \(n+1\).
---
## 3. Conclusion
By the principle of mathematical induction, the closed‐form expression
\[
\boxed{a_n = a \, r^n}
\]
holds for all integers \(n \ge 0\).
