solve the recurrence relation an+2 +3an+1 +2an=3^n , n>=0 ,a0=0, a1=1 . solve this using recursion

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.

---

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.

Learn how to solve the recurrence relation an+2 + 3an+1 + 2an = 3^n with initial conditions a0 = 0, a1 = 1. This step-by-step guide explains how to find both the homogeneous and particular solutions, and use recursion to solve for constants. Suitable for college-level students studying linear algebra or discrete mathematics.

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