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.