Math Problem Statement

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

Solution

The given recurrence relation is:

an=5an16an2a_{n} = 5a_{n-1} - 6a_{n-2}

with initial conditions a0=3a_0 = 3 and a1=4a_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 an=rna_n = r^n. Substituting into the recurrence relation gives:

rn=5rn16rn2r^n = 5r^{n-1} - 6r^{n-2}

Dividing by rn2r^{n-2} (assuming r0r \neq 0):

r2=5r6r^2 = 5r - 6

This simplifies to:

r25r+6=0r^2 - 5r + 6 = 0

Step 2: Solve the Characteristic Equation

We solve the quadratic equation:

r25r+6=0r^2 - 5r + 6 = 0

Factorizing:

(r2)(r3)=0(r - 2)(r - 3) = 0

So, the roots are:

r1=2,r2=3r_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:

an=C12n+C23na_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 C1C_1 and C2C_2.

For n=0n = 0:

a0=C120+C230=C1+C2=3a_0 = C_1 \cdot 2^0 + C_2 \cdot 3^0 = C_1 + C_2 = 3

For n=1n = 1:

a1=C121+C231=2C1+3C2=4a_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:

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$$. --- Would you like more details on any part of this process, or do you have any questions? --- Here are some related questions you might find interesting: 1. How would the solution change if the initial conditions were different? 2. What happens if the characteristic equation has complex roots? 3. How can you solve non-homogeneous recurrence relations? 4. What is the significance of the characteristic roots in determining the behavior of the sequence? 5. How would you approach solving a third-order linear recurrence relation? 6. Can you derive the formula for a sum of the sequence $$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.

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Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Characteristic Equation
Linear Combinations

Formulas

Quadratic equation solving
Linear combination formula

Theorems

Linear recurrence relations

Suitable Grade Level

Advanced High School