The solution of \(a_{n}=5 a_{n-1}-6 a_{n-2}\), where \(a_{0}=3, a_{1}=4\) is

The given recurrence relation is:

\[
a_{n} = 5a_{n-1} - 6a_{n-2}
\]

with initial conditions \(a_0 = 3\) and \(a_1 = 4\).

### Step 1: Find the Characteristic Equation
The characteristic equation corresponding to the recurrence relation is obtained by assuming a solution of the form \(a_n = r^n\). Substituting into the recurrence relation gives:

\[
r^n = 5r^{n-1} - 6r^{n-2}
\]

Dividing by \(r^{n-2}\) (assuming \(r \neq 0\)):

\[
r^2 = 5r - 6
\]

This simplifies to:

\[
r^2 - 5r + 6 = 0
\]

### Step 2: Solve the Characteristic Equation
We solve the quadratic equation:

\[
r^2 - 5r + 6 = 0
\]

Factorizing:

\[
(r - 2)(r - 3) = 0
\]

So, the roots are:

\[
r_1 = 2, \quad r_2 = 3
\]

### Step 3: General Solution of the Recurrence Relation
The general solution to the recurrence relation is a linear combination of the terms involving the roots:

\[
a_n = C_1 \cdot 2^n + C_2 \cdot 3^n
\]

### Step 4: Apply Initial Conditions
Now, use the initial conditions to find the constants \(C_1\) and \(C_2\).

For \(n = 0\):

\[
a_0 = C_1 \cdot 2^0 + C_2 \cdot 3^0 = C_1 + C_2 = 3
\]

For \(n = 1\):

\[
a_1 = C_1 \cdot 2^1 + C_2 \cdot 3^1 = 2C_1 + 3C_2 = 4
\]

### Step 5: Solve the System of Equations
We have the system of equations:

\[
\begin{aligned}
C_1 + C_2 &= 3 \quad \text{(1)} \\
2C_1 + 3C_2 &= 4 \quad \text{(2)}
\end{aligned}
\]

Using equation (1), express \(C_1\) as:

\[
C_1 = 3 - C_2
\]

Substitute this into equation (2):

\[
2(3 - C_2) + 3C_2 = 4
\]

Simplifying:

\[
6 - 2C_2 + 3C_2 = 4
\]

\[
C_2 = 2
\]

Now, substitute \(C_2 = 2\) into equation (1):

\[
C_1 + 2 = 3 \quad \Rightarrow \quad C_1 = 1
\]

### Final Solution
Thus, the solution to the recurrence relation is:

\[
a_n = 1 \cdot 2^n + 2 \cdot 3^n = 2^n + 2 \cdot 3^n
\]

This is the explicit formula for \(a_n\).

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Would you like more details on any part of this process, or do you have any questions?

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Here are some related questions you might find interesting:

1. How would the solution change if the initial conditions were different?
2. What happens if the characteristic equation has complex roots?
3. How can you solve non-homogeneous recurrence relations?
4. What is the significance of the characteristic roots in determining the behavior of the sequence?
5. How would you approach solving a third-order linear recurrence relation?
6. Can you derive the formula for a sum of the sequence \(a_n\)?
7. What are the applications of solving such recurrence relations in computer science?
8. How would you generalize this approach to solve a system of recurrence relations?

**Tip:** Always double-check your characteristic equation and roots, as errors there can lead to incorrect solutions in recurrence relations.

Explore how to solve the recurrence relation a_n = 5a_n-1 - 6a_n-2 with detailed step-by-step solutions. Understand the characteristic equation, linear combinations, and solving initial conditions. Suitable for advanced high school students and anyone interested in linear recurrence relations.

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