Solve the recurrence relation 𝑎௥ − 7𝑎௥ିଵ + 12𝑎௥ିଶ = 𝑟. 4
௥
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We are tasked with solving the recurrence relation:

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

where \( r \geq 4 \).

### Step 1: Solve the Homogeneous Equation

The first step is to solve the **homogeneous** part of the recurrence relation:

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

Assume a solution of the form \( a_r = x^r \). Substituting into the homogeneous equation:

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

Dividing through by \( x^{r-2} \) gives the **characteristic equation**:

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

We solve this quadratic equation:

\[
x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(12)}}{2(1)}
\]
\[
x = \frac{7 \pm \sqrt{49 - 48}}{2}
\]
\[
x = \frac{7 \pm 1}{2}
\]

So, the roots are:

\[
x = 4 \quad \text{and} \quad x = 3
\]

Thus, the general solution to the homogeneous equation is:

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

### Step 2: Solve the Non-Homogeneous Equation

Now, we solve the non-homogeneous equation:

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

We need to find a particular solution, \( a_r^{(p)} \), to this equation. Since the right-hand side is linear in \( r \), we try a particular solution of the form:

\[
a_r^{(p)} = Ar + B
\]

Substitute this into the recurrence relation:

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

Simplifying both sides:

\[
Ar + B - 7A(r - 1) - 7B + 12A(r - 2) + 12B = r
\]
\[
Ar + B - 7Ar + 7A - 7B + 12Ar - 24A + 12B = r
\]
\[
(Ar - 7Ar + 12Ar) + (B - 7B + 12B) + (7A - 24A) = r
\]
\[
6Ar + (6B - 17A) = r
\]

Equating the coefficients of \( r \) and the constant terms:

1. \( 6A = 1 \), which gives \( A = \frac{1}{6} \)
2. \( 6B - 17A = 0 \), which gives \( 6B - 17 \times \frac{1}{6} = 0 \), or \( 6B = \frac{17}{6} \), so \( B = \frac{17}{36} \)

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:

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

### Conclusion

The solution to the recurrence relation is:

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

To find the constants \( C_1 \) and \( C_2 \), you would need initial conditions (values for \( a_0 \), \( a_1 \), etc.).

---

Do you have any questions or would you like further details? Here are some questions that expand on the topic:

1. How do we handle initial conditions in recurrence relations?
2. Can we apply the same method for non-homogeneous terms of higher degree?
3. What happens if the characteristic equation has repeated roots?
4. How does this method apply to second-order linear recurrence relations in general?
5. How would you solve a recurrence relation using matrix methods?

**Tip:** Always check the form of the non-homogeneous term before assuming a particular solution!

Solve the recurrence relation 𝑎_r − 7𝑎_{r-1} + 12𝑎_{r-2} = r for r ≥ 4.

Explore how to solve the recurrence relation ar − 7ar-1 + 12ar-2 = r for r ≥ 4 with a detailed step-by-step solution. This problem covers essential concepts in recurrence relations, linear homogeneous equations, and the method of finding particular solutions. Ideal for Grades 11-12 or early undergraduate students.

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