Answer
\[
\boxed{
r_k \;=\; 1 \;+\; 3\,k,
\quad \text{for all integers } k \ge 0.
}
\]
Work Step by Step
We have the recurrence
\[
r_k \;=\; 2\,r_{k-1} \;-\; r_{k-2},
\quad \text{for all integers } k \ge 2,
\quad\text{with}\quad r_0 = 1,\; r_1 = 4.
\]
---
## 1. Solve the Characteristic Equation
First, we look for solutions of the form \(r_k = t^k\). Substituting into the recurrence:
\[
t^k = 2\,t^{k-1} \;-\; t^{k-2}.
\]
Divide both sides by \(t^{k-2}\) (assuming \(t\neq 0\)):
\[
t^2 = 2\,t - 1,
\]
which rearranges to
\[
t^2 - 2\,t + 1 = 0.
\]
Factor or recognize it as a perfect square:
\[
(t - 1)^2 = 0 \quad\Longrightarrow\quad t = 1 \quad(\text{a repeated root}).
\]
Hence, the characteristic equation has a **double root** at \(t=1\).
---
## 2. General Form for a Double Root
When the characteristic equation has a repeated root \(r\), the general solution to the homogeneous recurrence takes the form
\[
r_k = A \cdot r^k \;+\; B \,k\,r^k.
\]
Here, \(r=1\). Thus,
\[
r_k = A \cdot 1^k \;+\; B\,k\,1^k = A + B\,k.
\]
---
## 3. Use Initial Conditions to Find \(A\) and \(B\)
We have:
1. \(r_0 = A + B \cdot 0 = A = 1.\)
So \(A = 1.\)
2. \(r_1 = A + B \cdot 1 = 1 + B.\)
But \(r_1=4\). Hence
\[
1 + B = 4
\quad\Longrightarrow\quad
B = 3.
\]
Thus \(A=1\) and \(B=3\).
