Problem: 3- Solve the recurrence relation � �𝑟 −5𝑎𝑟−1 +6𝑎𝑟−2 = 2 for 𝑟 =2,3,4… and initial condition 𝑎0 = 1, and 𝑎1 = 1

The given recurrence relation is:

\[
a_r - 5a_{r-1} + 6a_{r-2} = 2
\]

with initial conditions:

\[
a_0 = 1, \quad a_1 = 1
\]

and for \( r = 2, 3, 4, \dots \).

### Step-by-step solution:

1. **Solve the homogeneous part:**
   The homogeneous part of the recurrence relation is:
   \[
   a_r - 5a_{r-1} + 6a_{r-2} = 0
   \]
   The characteristic equation for this recurrence is:
   \[
   x^2 - 5x + 6 = 0
   \]
   Factor this equation:
   \[
   (x - 2)(x - 3) = 0
   \]
   So, the roots are \( x = 2 \) and \( x = 3 \).

   The general solution to the homogeneous recurrence relation is:
   \[
   a_r^{(h)} = c_1 2^r + c_2 3^r
   \]
   where \( c_1 \) and \( c_2 \) are constants to be determined from the initial conditions.

2. **Solve the non-homogeneous part:**
   The non-homogeneous recurrence relation has a constant term on the right-hand side, so we look for a particular solution of the form \( a_r^{(p)} = A \), where \( A \) is a constant.

   Substitute \( a_r^{(p)} = A \) into the original recurrence:
   \[
   A - 5A + 6A = 2
   \]
   Simplifying:
   \[
   2A = 2 \quad \Rightarrow \quad A = 1
   \]

   Thus, the particular solution is \( a_r^{(p)} = 1 \).

3. **General solution:**
   The general solution to the recurrence relation is the sum of the homogeneous and particular solutions:
   \[
   a_r = a_r^{(h)} + a_r^{(p)} = c_1 2^r + c_2 3^r + 1
   \]

4. **Use the initial conditions:**
   Now, use the initial conditions \( a_0 = 1 \) and \( a_1 = 1 \) to determine \( c_1 \) and \( c_2 \).

   For \( r = 0 \):
   \[
   a_0 = c_1 2^0 + c_2 3^0 + 1 = c_1 + c_2 + 1 = 1
   \]
   This gives:
   \[
   c_1 + c_2 = 0 \quad \text{(Equation 1)}
   \]

   For \( r = 1 \):
   \[
   a_1 = c_1 2^1 + c_2 3^1 + 1 = 2c_1 + 3c_2 + 1 = 1
   \]
   Simplifying:
   \[
   2c_1 + 3c_2 = 0 \quad \text{(Equation 2)}
   \]

5. **Solve the system of equations:**
   From Equation 1: \( c_1 = -c_2 \).

   Substitute into Equation 2:
   \[
   2(-c_2) + 3c_2 = 0 \quad \Rightarrow \quad -2c_2 + 3c_2 = 0 \quad \Rightarrow \quad c_2 = 0
   \]

   Substituting \( c_2 = 0 \) into Equation 1:
   \[
   c_1 + 0 = 0 \quad \Rightarrow \quad c_1 = 0
   \]

6. **Final solution:**
   Therefore, the general solution to the recurrence relation is:
   \[
   a_r = 1
   \]

Thus, the solution is a constant sequence:

\[
a_r = 1 \quad \text{for all} \quad r \geq 0.
\]

Would you like further details or have any questions?

Here are five questions to expand on this:
1. What happens if we change the initial conditions?
2. How would the solution change if the recurrence relation were non-linear?
3. Can we solve the recurrence relation using matrix methods?
4. What is the significance of the characteristic equation?
5. How can recurrence relations be used in algorithm design?

**Tip:** Solving homogeneous and non-homogeneous parts separately is a common approach for linear recurrence relations.

Solve the recurrence relation a_r - 5a_{r-1} + 6a_{r-2} = 2 for r = 2, 3, 4... with initial conditions a_0 = 1 and a_1 = 1.

Explore how to solve the recurrence relation a_r - 5a_{r-1} + 6a_{r-2} = 2 with detailed steps. This problem covers solving linear homogeneous and non-homogeneous recurrence relations, characteristic equations, and using initial conditions. Suitable for students in Grades 10-12.

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