Solve the recurrence relation 𝑎௥ − 7𝑎௥ିଵ + 12𝑎௥ିଶ = 𝑟. 4
௥
.

The given recurrence relation is:

\[
a_r - 7a_{r-1} + 12a_{r-2} = r
\]

We can solve this non-homogeneous recurrence relation in two parts: the **homogeneous** solution and the **particular** solution.

### Step 1: Solve the homogeneous part

The homogeneous recurrence relation is:

\[
a_r - 7a_{r-1} + 12a_{r-2} = 0
\]

This is a second-order linear recurrence with constant coefficients. To solve it, we first find the characteristic equation by assuming a solution of the form \( a_r = \lambda^r \):

\[
\lambda^r - 7\lambda^{r-1} + 12\lambda^{r-2} = 0
\]

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

\[
\lambda^2 - 7\lambda + 12 = 0
\]

Now, solve this quadratic equation:

\[
\lambda^2 - 7\lambda + 12 = 0
\]

Factoring the quadratic:

\[
(\lambda - 3)(\lambda - 4) = 0
\]

Thus, the characteristic roots are \( \lambda_1 = 3 \) and \( \lambda_2 = 4 \).

The general solution to the homogeneous part is:

\[
a_r^{(h)} = C_1 \cdot 3^r + C_2 \cdot 4^r
\]

### Step 2: Solve the non-homogeneous part

Now, we find a particular solution for the non-homogeneous equation:

\[
a_r - 7a_{r-1} + 12a_{r-2} = r
\]

We try a particular solution of the form \( a_r^{(p)} = Ar + B \), where \( A \) and \( B \) are constants to be determined.

Substitute \( a_r^{(p)} = Ar + B \) into the recurrence relation:

\[
(Ar + B) - 7(A(r-1) + B) + 12(A(r-2) + B) = r
\]

Simplify the left-hand side:

\[
(Ar + B) - 7(A(r-1) + B) + 12(A(r-2) + B)
= Ar + B - 7A(r-1) - 7B + 12A(r-2) + 12B
\]

\[
= Ar + B - 7Ar + 7A - 7B + 12Ar - 24A + 12B
\]

\[
= (A - 7A + 12A)r + (B + 7A - 7B - 24A + 12B)
\]

\[
= 6Ar + (B - 7B + 12B + 7A - 24A)
\]

\[
= 6Ar + (6B - 17A)
\]

Equate this to the right-hand side of the original equation, which is \( r \):

\[
6Ar + (6B - 17A) = r
\]

From this, we get two equations by comparing the coefficients of \( r \) and the constant terms:

1. \( 6A = 1 \) ⟹ \( A = \frac{1}{6} \)
2. \( 6B - 17A = 0 \) ⟹ \( 6B - 17\left(\frac{1}{6}\right) = 0 \)

Solve for \( B \):

\[
6B - \frac{17}{6} = 0
\]

\[
6B = \frac{17}{6}
\]

\[
B = \frac{17}{36}
\]

Thus, the particular solution is:

\[
a_r^{(p)} = \frac{1}{6}r + \frac{17}{36}
\]

### Step 3: General solution

The general solution to the recurrence relation is the sum of the homogeneous and particular solutions:

\[
a_r = C_1 \cdot 3^r + C_2 \cdot 4^r + \frac{1}{6}r + \frac{17}{36}
\]

This is the complete solution to the recurrence relation.

Would you like more details or have any questions?

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### Relative Questions:
1. How do you find the characteristic equation of a recurrence relation?
2. What are the differences between homogeneous and non-homogeneous recurrence relations?
3. How can you verify a particular solution to a recurrence relation?
4. What is the significance of characteristic roots in solving recurrence relations?
5. How would the solution change if the right-hand side was a quadratic function instead of linear?

### Tip:
When solving non-homogeneous recurrence relations, always try a particular solution based on the form of the non-homogeneous term (e.g., constant, linear, quadratic, etc.).

Solve the recurrence relation 𝑎_r − 7𝑎_{r−1} + 12𝑎_{r−2} = r.

Discover how to solve the second-order recurrence relation a_r − 7a_{r−1} + 12a_{r−2} = r. This solution includes the homogeneous and non-homogeneous parts using the characteristic root method and the method of undetermined coefficients. Ideal for students in undergraduate discrete mathematics courses.

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