solve the recurrence relation an +an-1 -6an-2=0 for n=2, given that a0=-1 and a1=8

To solve the recurrence relation \( a_n + a_{n-1} - 6a_{n-2} = 0 \) for \( n = 2 \), with the initial conditions \( a_0 = -1 \) and \( a_1 = 8 \), we can proceed step by step.

### Step 1: Write out the recurrence relation.

The recurrence relation is:
\[
a_n + a_{n-1} - 6a_{n-2} = 0.
\]

We need to find \( a_2 \). To do this, we will use the given initial conditions.

### Step 2: Use the recurrence for \( n = 2 \).

Substitute \( n = 2 \) into the recurrence relation:
\[
a_2 + a_1 - 6a_0 = 0.
\]

### Step 3: Substitute the values of \( a_0 \) and \( a_1 \).

From the given initial conditions:
- \( a_0 = -1 \),
- \( a_1 = 8 \).

Substituting these into the equation:
\[
a_2 + 8 - 6(-1) = 0.
\]

Simplifying:
\[
a_2 + 8 + 6 = 0,
\]
\[
a_2 + 14 = 0,
\]
\[
a_2 = -14.
\]

### Final Answer:
The value of \( a_2 \) is \( -14 \).

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Would you like a detailed explanation of recurrence relations or a more general approach to solving them?

Here are 5 related questions:
1. How do you solve recurrence relations with constant coefficients?
2. What is the general solution form for a second-order linear recurrence relation?
3. How can you determine the characteristic equation of a recurrence relation?
4. What is the role of initial conditions in solving recurrence relations?
5. How do you solve recurrence relations using the method of generating functions?

**Tip**: When solving recurrence relations, always check if the relation is linear and homogeneous first, as this simplifies the solution process.

Learn how to solve the recurrence relation an + an-1 - 6an-2 = 0 for n=2 with given initial conditions a0 = -1 and a1 = 8. The solution uses a step-by-step approach to calculate a2. Suitable for high school students in Grades 11-12.

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