Solve the recurrence relation a_n = -3a_(n-1) - 3a_(n-2) - a_(n-3) for n ≥ 2, a_0 = 5, a_1 = -9, and a_2 = 15

To solve the recurrence relation 

\[
a_n = -3a_{n-1} - 3a_{n-2} - a_{n-3} \quad \text{for } n \geq 2
\]

with the initial conditions \( a_0 = 5 \), \( a_1 = -9 \), and \( a_2 = 15 \), we proceed as follows:

### Step 1: Set up the characteristic equation
The associated characteristic equation for the recurrence relation is found by assuming a solution of the form \( a_n = r^n \). Substituting this into the recurrence relation gives:

\[
r^n = -3r^{n-1} - 3r^{n-2} - r^{n-3}
\]

Dividing through by \( r^{n-3} \) (assuming \( r \neq 0 \)):

\[
r^3 + 3r^2 + 3r + 1 = 0
\]

### Step 2: Solve the characteristic equation
The characteristic equation can be factored as follows:

\[
r^3 + 3r^2 + 3r + 1 = (r + 1)^3 = 0
\]

So, the characteristic equation has a repeated root \( r = -1 \) with multiplicity 3.

### Step 3: Write the general solution
Since \( r = -1 \) is a repeated root with multiplicity 3, the general solution to the recurrence relation is:

\[
a_n = (C_1 + C_2n + C_3n^2)(-1)^n
\]

where \( C_1 \), \( C_2 \), and \( C_3 \) are constants to be determined from the initial conditions.

### Step 4: Apply the initial conditions
Now, we use the initial conditions to determine \( C_1 \), \( C_2 \), and \( C_3 \).

1. For \( n = 0 \):
   \[
   a_0 = C_1(-1)^0 = C_1 = 5
   \]
   So, \( C_1 = 5 \).

2. For \( n = 1 \):
   \[
   a_1 = (C_1 + C_2 \cdot 1 + C_3 \cdot 1^2)(-1)^1 = -(C_1 + C_2 + C_3)
   \]
   Given \( a_1 = -9 \):
   \[
   -(5 + C_2 + C_3) = -9 \implies 5 + C_2 + C_3 = 9 \implies C_2 + C_3 = 4
   \]

3. For \( n = 2 \):
   \[
   a_2 = (C_1 + C_2 \cdot 2 + C_3 \cdot 2^2)(-1)^2 = C_1 + 2C_2 + 4C_3
   \]
   Given \( a_2 = 15 \):
   \[
   5 + 2C_2 + 4C_3 = 15 \implies 2C_2 + 4C_3 = 10 \implies C_2 + 2C_3 = 5
   \]

### Step 5: Solve for \( C_2 \) and \( C_3 \)
We now have the system of linear equations:
\[
C_2 + C_3 = 4
\]
\[
C_2 + 2C_3 = 5
\]

Subtract the first equation from the second:
\[
(C_2 + 2C_3) - (C_2 + C_3) = 5 - 4 \implies C_3 = 1
\]

Substitute \( C_3 = 1 \) into the first equation:
\[
C_2 + 1 = 4 \implies C_2 = 3
\]

### Final Answer
The general solution is:

\[
a_n = (5 + 3n + n^2)(-1)^n
\]

This is the closed-form expression for the sequence.

Would you like more details, or do you have any questions?

### Related Questions
1. How would you solve a recurrence relation with complex roots in the characteristic equation?
2. What is the impact of different initial conditions on the solution to a recurrence relation?
3. How does the multiplicity of roots affect the general solution of a recurrence relation?
4. Can the method used here be extended to non-homogeneous recurrence relations?
5. How would you approach solving a non-linear recurrence relation?

### Tip
When solving recurrence relations, always check the multiplicity of the roots of the characteristic equation, as it significantly affects the structure of the general solution.

Learn how to solve the recurrence relation a_n = -3a_(n-1) - 3a_(n-2) - a_(n-3) with initial conditions a_0 = 5, a_1 = -9, and a_2 = 15. This solution involves finding the characteristic equation and handling repeated roots, ideal for undergraduate students studying recurrence relations.

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