To solve the recurrence relation $a_{r+2} - 5a_{r+1} + 6a_r = 2$ using the method of generating functions, we follow these steps:

1. Define the Generating Function

Let the generating function be: $A(x) = \sum_{r=0}^\infty a_r x^r.$

2. Write the Recurrence Relation in Terms of Generating Functions

Multiply the recurrence relation $a_{r+2} - 5a_{r+1} + 6a_r = 2$ by $x^r$ and sum over $r \geq 0$: $\sum_{r=0}^\infty \left( a_{r+2}x^r - 5a_{r+1}x^r + 6a_r x^r \right) = \sum_{r=0}^\infty 2x^r.$

3. Shift the Indices to Align Terms

• For $\sum_{r=0}^\infty a_{r+2}x^r$, substitute $r \to r-2$: $\sum_{r=0}^\infty a_{r+2}x^r = x^2 \sum_{r=0}^\infty a_r x^r = x^2 A(x).$
• For $\sum_{r=0}^\infty a_{r+1}x^r$, substitute $r \to r-1$: $\sum_{r=0}^\infty a_{r+1}x^r = x \sum_{r=0}^\infty a_r x^r = x A(x).$
• The term $\sum_{r=0}^\infty a_r x^r = A(x)$ remains unchanged.

Substitute these shifts into the recurrence: $x^2 A(x) - 5x A(x) + 6A(x) = \frac{2}{1-x}.$

4. Factorize and Solve for $A(x)$

Group $A(x)$: $A(x) \left( x^2 - 5x + 6 \right) = \frac{2}{1-x}.$ Factorize $x^2 - 5x + 6 = (x-2)(x-3)$: $A(x) = \frac{2}{(1-x)(x-2)(x-3)}.$

5. Partial Fraction Decomposition

Decompose $\frac{2}{(1-x)(x-2)(x-3)}$: $\frac{2}{(1-x)(x-2)(x-3)} = \frac{A}{1-x} + \frac{B}{x-2} + \frac{C}{x-3}.$ Multiply through by $(1-x)(x-2)(x-3)$ and solve for $A, B, C$: $2 = A(x-2)(x-3) + B(1-x)(x-3) + C(1-x)(x-2).$ Expanding and equating coefficients gives: $A = -1, \, B = 3, \, C = 0.$

Thus: $A(x) = \frac{-1}{1-x} + \frac{3}{x-2}.$

6. Expand into Power Series

Expand each term:

• $\frac{-1}{1-x} = -\sum_{r=0}^\infty x^r$.
• ( \frac{3}{x-2