To solve the recurrence relation:

$a_{n+2} + 3a_{n+1} + 2a_n = 3^n \quad \text{for} \quad n \geq 0$ with initial conditions $a_0 = 0$ and $a_1 = 1$, we will break this down into steps using recursion.

Step 1: Solve the Homogeneous Equation

First, solve the corresponding homogeneous recurrence relation: $a_{n+2} + 3a_{n+1} + 2a_n = 0$

This is a second-order linear recurrence relation. The characteristic equation for this is: $r^2 + 3r + 2 = 0$

Factoring the quadratic equation: $(r + 1)(r + 2) = 0$

The roots are $r = -1$ and $r = -2$.

Thus, the general solution to the homogeneous equation is: $a_n^{(h)} = C_1 (-1)^n + C_2 (-2)^n$ where $C_1$ and $C_2$ are constants to be determined.

Step 2: Solve the Particular Solution

Now, we solve the non-homogeneous recurrence relation: $a_{n+2} + 3a_{n+1} + 2a_n = 3^n$

We propose a particular solution of the form: $a_n^{(p)} = A \cdot 3^n$ Substitute this into the recurrence relation: $A \cdot 3^{n+2} + 3A \cdot 3^{n+1} + 2A \cdot 3^n = 3^n$ Factor out $3^n$: $A \cdot 9 + 3A \cdot 3 + 2A = 1$ Simplifying: $9A + 9A + 2A = 1 \quad \Rightarrow \quad 20A = 1 \quad \Rightarrow \quad A = \frac{1}{20}$

Thus, the particular solution is: $a_n^{(p)} = \frac{1}{20} \cdot 3^n$

Step 3: General Solution

The general solution to the non-homogeneous recurrence relation is the sum of the homogeneous and particular solutions: $a_n = C_1 (-1)^n + C_2 (-2)^n + \frac{1}{20} \cdot 3^n$

Step 4: Use Initial Conditions

We use the initial conditions $a_0 = 0$ and $a_1 = 1$ to find $C_1$ and $C_2$.

1. At $n = 0$: $a_0 = C_1 (-1)^0 + C_2 (-2)^0 + \frac{1}{20} \cdot 3^0 = C_1 + C_2 + \frac{1}{20} = 0$ $C_1 + C_2 = -\frac{1}{20}$

2. At $n = 1$: $a_1 = C_1 (-1)^1 + C_2 (-2)^1 + \frac{1}{20} \cdot 3^1 = -C_1 - 2C_2 + \frac{3}{20} = 1$ $-C_1 - 2C_2 + \frac{3}{20} = 1$ $-C_1 - 2C_2 = 1 - \frac{3}{20} = \frac{17}{20}$

Now, solve the system of equations: $C_1 + C_2 = -\frac{1}{20}$ $-C_1 - 2C_2 = \frac{17}{20}$

From the first equation, $C_1 = -\frac{1}{20} - C_2$. Substitute this into the second equation: $-\left( -\frac{1}{20} - C_2 \right) - 2C_2 = \frac{17}{20}$ $\frac{1}{20} + C_2 - 2C_2 = \frac{17}{20}$ $\frac{1}{20} - C_2 = \frac{17}{20}$ $-C_2 = \frac{17}{20} - \frac{1}{20} = \frac{16}{20} = \frac{4}{5}$ $C_2 = -\frac{4}{5}$

Substitute $C_2 = -\frac{4}{5}$ into $C_1 + C_2 = -\frac{1}{20}$: $C_1 - \frac{4}{5} = -\frac{1}{20}$ $C_1 = -\frac{1}{20} + \frac{4}{5} = -\frac{1}{20} + \frac{16}{20} = \frac{15}{20} = \frac{3}{4}$

Step 5: Final Solution

Thus, the final solution to the recurrence relation is: $a_n = \frac{3}{4} (-1)^n - \frac{4}{5} (-2)^n + \frac{1}{20} \cdot 3^n$

This is the explicit form of the solution to the recurrence relation.

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Do you have any questions or need further details on any step?

Here are 5 related questions to expand on this topic:

1. How do we solve higher-order recurrence relations?
2. What happens if the characteristic equation has repeated roots?
3. Can you derive the recurrence relation for Fibonacci numbers?
4. How do non-homogeneous terms affect the solution structure?
5. Can generating functions be used to solve recurrence relations?

Tip: Always check the characteristic equation carefully for potential complex or repeated roots, as they affect the form of the solution.