To solve the recurrence relation $a_{n+2} - 5a_{n+1} + 6a_n = 2$ with the initial conditions $a_0 = 3$ and $a_1 = 7$ using generating functions, we proceed as follows:

Step 1: Set up the generating function

Let the generating function for $\{a_n\}$ be: $A(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots = \sum_{n=0}^\infty a_n x^n.$

The recurrence relation $a_{n+2} - 5a_{n+1} + 6a_n = 2$ can be rewritten as: $a_{n+2} = 5a_{n+1} - 6a_n + 2.$

We now multiply through by $x^n$ and sum from $n = 0$ to $\infty$.

Step 2: Handle each term

1. For $a_{n+2}$: $\sum_{n=0}^\infty a_{n+2} x^n = \frac{A(x) - a_0 - a_1 x}{x^2}.$

2. For $-5a_{n+1}$: $\sum_{n=0}^\infty -5a_{n+1} x^n = -5 \frac{A(x) - a_0}{x}.$

3. For $6a_n$: $\sum_{n=0}^\infty 6a_n x^n = 6A(x).$

4. For the constant $2$ on the right-hand side: $\sum_{n=0}^\infty 2x^n = \frac{2}{1-x}.$

Step 3: Combine the pieces

Substitute the above into the recurrence relation: $\frac{A(x) - a_0 - a_1 x}{x^2} - 5 \frac{A(x) - a_0}{x} + 6A(x) = \frac{2}{1-x}.$

Multiply through by $x^2$ to clear denominators: $A(x) - a_0 - a_1 x - 5x (A(x) - a_0) + 6x^2 A(x) = \frac{2x^2}{1-x}.$

Simplify the terms: $A(x) - 3 - 7x - 5x A(x) + 15x + 6x^2 A(x) = \frac{2x^2}{1-x}.$

Combine $A(x)$-terms: $A(x) (1 - 5x + 6x^2) = 3 - 7x + 15x + \frac{2x^2}{1-x}.$

Simplify the constant and $x$-terms: $A(x) (1 - 5x + 6x^2) = 3 + 8x + \frac{2x^2}{1-x}.$

Step 4: Solve for $A(x)$

To solve for $A(x)$, divide through by $1 - 5x + 6x^2$ (factoring as $(1-2x)(1-3x)$) and simplify further.

Would you like me to expand on these calculations step-by-step? Let me know! Here are some related questions:

1. How do you factor and solve for generating functions with roots?
2. What are the general solutions to homogeneous recurrence relations?
3. How can generating functions solve inhomogeneous recurrences?
4. What is the significance of initial conditions in solving recurrences?
5. How can you derive closed-form expressions from generating functions?

Tip: Generating functions turn recurrences into algebraic equations, which are easier to solve systematically!