The given recurrence relation is:

$a_r - 5a_{r-1} + 6a_{r-2} = 2$

with initial conditions:

$a_0 = 1, \quad a_1 = 1$

and for $r = 2, 3, 4, \dots$.

Step-by-step solution:

1. Solve the homogeneous part: The homogeneous part of the recurrence relation is: $a_r - 5a_{r-1} + 6a_{r-2} = 0$ The characteristic equation for this recurrence is: $x^2 - 5x + 6 = 0$ Factor this equation: $(x - 2)(x - 3) = 0$ So, the roots are $x = 2$ and $x = 3$.

   The general solution to the homogeneous recurrence relation is: $a_r^{(h)} = c_1 2^r + c_2 3^r$ where $c_1$ and $c_2$ are constants to be determined from the initial conditions.

2. Solve the non-homogeneous part: The non-homogeneous recurrence relation has a constant term on the right-hand side, so we look for a particular solution of the form $a_r^{(p)} = A$, where $A$ is a constant.

   Substitute $a_r^{(p)} = A$ into the original recurrence: $A - 5A + 6A = 2$ Simplifying: $2A = 2 \quad \Rightarrow \quad A = 1$

   Thus, the particular solution is $a_r^{(p)} = 1$.

3. General solution: The general solution to the recurrence relation is the sum of the homogeneous and particular solutions: $a_r = a_r^{(h)} + a_r^{(p)} = c_1 2^r + c_2 3^r + 1$

4. Use the initial conditions: Now, use the initial conditions $a_0 = 1$ and $a_1 = 1$ to determine $c_1$ and $c_2$.

   For $r = 0$: $a_0 = c_1 2^0 + c_2 3^0 + 1 = c_1 + c_2 + 1 = 1$ This gives: $c_1 + c_2 = 0 \quad \text{(Equation 1)}$

   For $r = 1$: $a_1 = c_1 2^1 + c_2 3^1 + 1 = 2c_1 + 3c_2 + 1 = 1$ Simplifying: $2c_1 + 3c_2 = 0 \quad \text{(Equation 2)}$

5. Solve the system of equations: From Equation 1: $c_1 = -c_2$.

   Substitute into Equation 2: $2(-c_2) + 3c_2 = 0 \quad \Rightarrow \quad -2c_2 + 3c_2 = 0 \quad \Rightarrow \quad c_2 = 0$

   Substituting $c_2 = 0$ into Equation 1: $c_1 + 0 = 0 \quad \Rightarrow \quad c_1 = 0$

6. Final solution: Therefore, the general solution to the recurrence relation is: $a_r = 1$

Thus, the solution is a constant sequence:

$a_r = 1 \quad \text{for all} \quad r \geq 0.$

Would you like further details or have any questions?

Here are five questions to expand on this:

1. What happens if we change the initial conditions?
2. How would the solution change if the recurrence relation were non-linear?
3. Can we solve the recurrence relation using matrix methods?
4. What is the significance of the characteristic equation?
5. How can recurrence relations be used in algorithm design?

Tip: Solving homogeneous and non-homogeneous parts separately is a common approach for linear recurrence relations.