Using the method of characteristic equation find solution for recurrence 
relation 
(i) 𝑎௡ = 3 𝑎௡ିଵ − 3𝑎௡ିଶ + 𝑎௡ିଷ , 𝑎଴ = 0, 𝑎ଷ = 3, 𝑎ହ = 10

To solve the recurrence relation:

\[
a_n = 3a_{n-1} - 3a_{n-2} + a_{n-3}
\]

using the **method of characteristic equations**, follow these steps:

### Step 1: Set up the characteristic equation

Assume the solution is 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 the entire equation by \( r^{n-3} \), we get the characteristic equation:

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

### Step 2: Solve the characteristic equation

Rearrange the characteristic equation:

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

Notice that this is a perfect cube, which can be factored as:

\[
(r - 1)^3 = 0
\]

Thus, the only root of the characteristic equation is \( r = 1 \) with multiplicity 3.

### Step 3: General solution form

Since \( r = 1 \) is a repeated root with multiplicity 3, the general solution to the recurrence relation is of the form:

\[
a_n = (c_1 + c_2n + c_3n^2)1^n
\]

This simplifies to:

\[
a_n = c_1 + c_2n + c_3n^2
\]

### Step 4: Use initial conditions to find the constants

We are given three initial conditions: \( a_0 = 0 \), \( a_3 = 3 \), and \( a_5 = 10 \).

#### Using \( a_0 = 0 \):

\[
a_0 = c_1 + c_2 \cdot 0 + c_3 \cdot 0^2 = c_1
\]
Thus, \( c_1 = 0 \).

#### Using \( a_3 = 3 \):

\[
a_3 = c_1 + c_2 \cdot 3 + c_3 \cdot 3^2 = 0 + 3c_2 + 9c_3 = 3
\]
This gives the equation:

\[
3c_2 + 9c_3 = 3
\]

Simplifying:

\[
c_2 + 3c_3 = 1 \quad \text{(Equation 1)}
\]

#### Using \( a_5 = 10 \):

\[
a_5 = c_1 + c_2 \cdot 5 + c_3 \cdot 5^2 = 0 + 5c_2 + 25c_3 = 10
\]
This gives the equation:

\[
5c_2 + 25c_3 = 10
\]

Simplifying:

\[
c_2 + 5c_3 = 2 \quad \text{(Equation 2)}
\]

### Step 5: Solve the system of equations

We have the following system of linear equations:

1. \( c_2 + 3c_3 = 1 \)
2. \( c_2 + 5c_3 = 2 \)

Subtract Equation 1 from Equation 2:

\[
(c_2 + 5c_3) - (c_2 + 3c_3) = 2 - 1
\]
\[
2c_3 = 1
\]
\[
c_3 = \frac{1}{2}
\]

Substitute \( c_3 = \frac{1}{2} \) into Equation 1:

\[
c_2 + 3 \cdot \frac{1}{2} = 1
\]
\[
c_2 + \frac{3}{2} = 1
\]
\[
c_2 = 1 - \frac{3}{2} = -\frac{1}{2}
\]

### Step 6: Final solution

Now that we have \( c_1 = 0 \), \( c_2 = -\frac{1}{2} \), and \( c_3 = \frac{1}{2} \), the general solution is:

\[
a_n = -\frac{1}{2}n + \frac{1}{2}n^2
\]

Simplifying:

\[
a_n = \frac{1}{2}n(n - 1)
\]

This is the solution to the recurrence relation.

---

Would you like further details on any step, or do you have any questions? 

Here are 5 related questions you might find helpful:

1. How would the solution change if the characteristic equation had distinct roots?
2. What is the general approach for solving non-homogeneous recurrence relations?
3. Can we solve recurrence relations without initial conditions? If so, how?
4. How would you apply the method of undetermined coefficients for different types of sequences?
5. What are the differences between linear and non-linear recurrence relations?

**Tip:** For recurrence relations with repeated roots, always include higher powers of \( n \) in the general solution to account for the multiplicity.

Using the method of characteristic equation find solution for recurrence relation (i) 𝑎௡ = 3 𝑎௡ିଵ − 3𝑎௡ିଶ + 𝑎௡ିଷ , 𝑎଴ = 0, 𝑎ଷ = 3, 𝑎ହ = 10

Discover how to solve the recurrence relation an = 3an-1 - 3an-2 + an-3 using the method of characteristic equations. This step-by-step guide covers solving the characteristic equation, handling repeated roots, and applying initial conditions a0 = 0, a3 = 3, a5 = 10 to find the general solution. Suitable for advanced high school and undergraduate students.

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