To solve the recurrence relation using generating functions, we proceed step by step.

Problem Statement:

The recurrence relation is:

$a_{n+2} - 5a_{n+1} + 6a_n = 2, \quad a_0 = 3, \quad a_1 = 7$

Step 1: Define the Generating Function

Let $A(x)$ be the generating function of the sequence $a_n$, i.e.,

$A(x) = \sum_{n=0}^{\infty} a_n x^n$

We aim to express the recurrence relation in terms of $A(x)$ and solve for $A(x)$.

Step 2: Express the Recurrence Using the Generating Function

Now, consider the generating functions for each term in the recurrence relation:

• The term $a_{n+2}$ corresponds to shifting the sequence by 2. Its generating function is $A(x) - a_0 - a_1 x$, because: $\sum_{n=2}^{\infty} a_{n+2} x^n = x^2 A(x)$
• The term $a_{n+1}$ corresponds to a shift by 1. Its generating function is $A(x) - a_0$, because: $\sum_{n=1}^{\infty} a_{n+1} x^n = x A(x)$
• The term $a_n$ remains the same and corresponds directly to $A(x)$.

So, the recurrence relation becomes:

$(x^2 A(x) - a_0 - a_1 x) - 5(x A(x) - a_0) + 6A(x) = \frac{2}{1 - x}$

where $\frac{2}{1 - x}$ is the generating function for the constant sequence 2 on the right-hand side.

Step 3: Substitute Initial Conditions

Substitute $a_0 = 3$ and $a_1 = 7$:

$(x^2 A(x) - 3 - 7x) - 5(x A(x) - 3) + 6A(x) = \frac{2}{1 - x}$

Now, expand and simplify:

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

Group terms involving $A(x)$:

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

Simplify the right-hand side:

$(x^2 - 5x + 6) A(x) = \frac{2}{1 - x} - 18 + 7x$

Step 4: Solve for $A(x)$

Simplify the equation:

$A(x) = \frac{2 - 18(1 - x) + 7x(1 - x)}{(x^2 - 5x + 6)(1 - x)}$

We will then expand the expression and use partial fraction decomposition to obtain $A(x)$. After this, we can expand $A(x)$ as a power series to find the closed form of the sequence $a_n$.

Would you like me to carry out the partial fraction decomposition and derive the solution?