Answer
See explanation
Work Step by Step
**Solution Explanation**
We are given the explicit formula for the sequence:
\[
a_n = 3n + 1 \quad \text{for all integers } n \ge 0,
\]
and we want to show it satisfies the recurrence relation
\[
a_k = a_{k-1} + 3 \quad \text{for all integers } k \ge 1.
\]
---
### Step 1. Express \(a_k\) and \(a_{k-1}\) from the Given Formula
Using \(a_n = 3n + 1\):
\(a_k = 3k + 1.\)
\(a_{k-1} = 3(k-1) + 1 = 3k - 3 + 1 = 3k - 2.\)
---
### Step 2. Verify the Recurrence
We check whether \(a_k\) equals \(a_{k-1} + 3\). Compute:
\[
a_{k-1} + 3 = (3k - 2) + 3 = 3k + 1,
\]
which is exactly \(a_k\).
Hence,
\[
a_k = a_{k-1} + 3 \quad \text{for all } k \ge 1,
\]
as required.
