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.