find the solutionb) an = 2an - 1 + 5 -2-6an-3 for n-3, 4, 5....., with initial conditions a0= 7 . a1 = - 4 ,and a2=8 

To solve the recurrence relation given by:

\[
a_n = 2a_{n-1} + 5 - 2 - 6a_{n-3} \quad \text{for } n = 3, 4, 5, \ldots
\]

with the initial conditions \(a_0 = 7\), \(a_1 = -4\), and \(a_2 = 8\), let's first simplify the given relation.

### Step 1: Simplify the recurrence relation

Simplify the expression inside the relation:

\[
a_n = 2a_{n-1} + 5 - 2 - 6a_{n-3} 
\]

Combine like terms:

\[
a_n = 2a_{n-1} + 3 - 6a_{n-3}
\]

So the recurrence relation becomes:

\[
a_n = 2a_{n-1} + 3 - 6a_{n-3}
\]

### Step 2: Compute the first few terms

We can use the initial conditions \(a_0 = 7\), \(a_1 = -4\), and \(a_2 = 8\) to find the next few terms.

1. **For \(n = 3\):**

\[
a_3 = 2a_2 + 3 - 6a_0
\]

Substitute the known values:

\[
a_3 = 2(8) + 3 - 6(7) = 16 + 3 - 42 = -23
\]

2. **For \(n = 4\):**

\[
a_4 = 2a_3 + 3 - 6a_1
\]

Substitute the values:

\[
a_4 = 2(-23) + 3 - 6(-4) = -46 + 3 + 24 = -19
\]

3. **For \(n = 5\):**

\[
a_5 = 2a_4 + 3 - 6a_2
\]

Substitute the values:

\[
a_5 = 2(-19) + 3 - 6(8) = -38 + 3 - 48 = -83
\]

### Step 3: Solution Summary

The values of \(a_n\) for \(n = 3, 4, 5\) are:

- \(a_3 = -23\)
- \(a_4 = -19\)
- \(a_5 = -83\)

These values follow the recurrence relation given the initial conditions.

### Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do we find a general formula for this type of recurrence relation?
2. What is the characteristic equation associated with this recurrence?
3. How do initial conditions affect the solution to a recurrence relation?
4. Can you solve a recurrence relation using a generating function?
5. What are the differences between homogeneous and non-homogeneous recurrence relations?

**Tip:** When dealing with recurrence relations, always check if you can simplify the given expression first to make calculations easier.

Find the solution for a_n = 2a_{n-1} + 5 - 2 - 6a_{n-3} for n = 3, 4, 5..., with initial conditions a0 = 7, a1 = -4, and a2 = 8.

Explore the step-by-step solution to the recurrence relation a_n = 2a_{n-1} + 3 - 6a_{n-3}, with initial conditions a0 = 7, a1 = -4, and a2 = 8. Understand how to calculate the next terms in the sequence for n = 3, 4, and 5. Ideal for high school students in Grades 10-12 studying recurrence relations.

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