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
\[
\begin{cases}
a_0 \text{ (given initial term)}, \\
a_k = a_{k-1} + d \quad \text{for all integers } k \ge 1,
\end{cases}
\]
has the closed‐form expression
\[
a_n = a_0 + nd.
\]
---
## Step 1. Base Case \((n = 0)\)
Check the formula for \(n = 0\):
\[
a_0 \stackrel{?}{=} a_0 + 0 \cdot d.
\]
Indeed,
\[
a_0 = a_0 + 0 = a_0.
\]
So the statement holds for \(n = 0\).
---
## Step 2. Inductive Step
Assume that for some integer \(n \ge 0\), the formula holds:
\[
a_n = a_0 + nd.
\]
We want to prove that it then holds for \(n + 1\). By the recursive definition:
\[
a_{n+1} = a_n + d.
\]
Using the inductive hypothesis \(a_n = a_0 + nd\), substitute:
\[
a_{n+1}
= \bigl(a_0 + nd\bigr) + d
= a_0 + (n+1)d.
\]
Thus, if the formula is true for \(n\), it is also true for \(n+1\).
---
## Step 3. Conclusion
By the principle of mathematical induction, the closed‐form expression
\[
\boxed{a_n = a_0 + nd}
\]
holds for all integers \(n \ge 0\).
