find the recurrence relation of following function- an+2 -3 a n+1 -2an=3" where-a0 = 0, a1=1

The given recurrence relation is:

\[
a_{n+2} - 3a_{n+1} - 2a_n = 3,
\]

with initial conditions \( a_0 = 0 \) and \( a_1 = 1 \).

### Step 1: Solve the homogeneous recurrence relation
The homogeneous part of the recurrence is:

\[
a_{n+2} - 3a_{n+1} - 2a_n = 0.
\]

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

\[
r^{n+2} - 3r^{n+1} - 2r^n = 0.
\]

Divide through by \( r^n \) (for \( r \neq 0 \)):

\[
r^2 - 3r - 2 = 0.
\]

Factoring:

\[
(r - 4)(r + 1) = 0.
\]

Thus, the roots are \( r = 4 \) and \( r = -1 \). The general solution of the homogeneous recurrence is:

\[
a_n^{(h)} = C_1(4^n) + C_2(-1)^n,
\]

where \( C_1 \) and \( C_2 \) are constants to be determined.

### Step 2: Solve the non-homogeneous recurrence relation
The non-homogeneous equation is:

\[
a_{n+2} - 3a_{n+1} - 2a_n = 3.
\]

Assume a particular solution \( a_n^{(p)} \) is constant. Let \( a_n^{(p)} = A \). Substituting \( a_n^{(p)} = A \) into the equation:

\[
A - 3A - 2A = 3.
\]

Simplify:

\[
-4A = 3 \quad \implies \quad A = -\frac{3}{4}.
\]

Thus, the particular solution is:

\[
a_n^{(p)} = -\frac{3}{4}.
\]

### Step 3: General solution
The general solution is the sum of the homogeneous and particular solutions:

\[
a_n = C_1(4^n) + C_2(-1)^n - \frac{3}{4}.
\]

### Step 4: Apply initial conditions
Using \( a_0 = 0 \) and \( a_1 = 1 \):

1. For \( n = 0 \), \( a_0 = 0 \):
   \[
   C_1(4^0) + C_2(-1)^0 - \frac{3}{4} = 0.
   \]
   Simplify:
   \[
   C_1 + C_2 - \frac{3}{4} = 0.
   \]
   \[
   C_1 + C_2 = \frac{3}{4}. \tag{1}
   \]

2. For \( n = 1 \), \( a_1 = 1 \):
   \[
   C_1(4^1) + C_2(-1)^1 - \frac{3}{4} = 1.
   \]
   Simplify:
   \[
   4C_1 - C_2 - \frac{3}{4} = 1.
   \]
   \[
   4C_1 - C_2 = \frac{7}{4}. \tag{2}
   \]

Solve the system of equations \((1)\) and \((2)\):
- Add equations \((1)\) and \((2)\):
  \[
  C_1 + C_2 + 4C_1 - C_2 = \frac{3}{4} + \frac{7}{4}.
  \]
  \[
  5C_1 = \frac{10}{4}.
  \]
  \[
  C_1 = \frac{1}{2}.
  \]

- Substitute \( C_1 = \frac{1}{2} \) into \((1)\):
  \[
  \frac{1}{2} + C_2 = \frac{3}{4}.
  \]
  \[
  C_2 = \frac{1}{4}.
  \]

### Final Solution
The solution is:

\[
a_n = \frac{1}{2}(4^n) + \frac{1}{4}(-1)^n - \frac{3}{4}.
\]

### Want more details or have questions?  
#### Related questions:
1. How can we confirm the correctness of the solution by induction?
2. What happens if the non-homogeneous term is a polynomial instead of a constant?
3. Can we generalize the solution for different initial conditions?
4. What role does the characteristic equation play in finding solutions?
5. How do we handle cases where the roots of the characteristic equation are repeated?

#### Tip:
Always verify the solution by substituting it back into the original recurrence relation!

Find the recurrence relation of the following function: a_{n+2} - 3a_{n+1} - 2a_n = 3 where a_0 = 0, a_1 = 1.

Learn how to solve the recurrence relation a_{n+2} - 3a_{n+1} - 2a_n = 3 with initial conditions a_0 = 0 and a_1 = 1. Explore the step-by-step process, including solving the characteristic equation and applying initial conditions. Ideal for Grades 10-12 students.

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