The given recurrence relation is:

$a_{n+2} + 4a_{n-1} + 4a_0 = 7 \quad \text{for} \quad n \geq 0$

with the initial conditions:

$a_0 = 1, \quad a_1 = 2$

To solve this, let's analyze the equation step by step.

Step 1: Plug in the initial condition $a_0 = 1$.

$a_{n+2} + 4a_{n-1} + 4(1) = 7$ Simplifying the equation:

$a_{n+2} + 4a_{n-1} + 4 = 7$ $a_{n+2} + 4a_{n-1} = 3$

This is the simplified recurrence relation:

$a_{n+2} = 3 - 4a_{n-1}$

Step 2: Use initial conditions to compute the first few terms.

We have $a_0 = 1$ and $a_1 = 2$. Let's find $a_2$ and $a_3$ using the recurrence relation.

For $n = 0$: $a_2 = 3 - 4a_0 = 3 - 4(1) = 3 - 4 = -1$

For $n = 1$: $a_3 = 3 - 4a_1 = 3 - 4(2) = 3 - 8 = -5$

Step 3: Verify the pattern and calculate more terms if needed.

We now have:

$a_0 = 1, \quad a_1 = 2, \quad a_2 = -1, \quad a_3 = -5$

We could continue computing more terms using the recurrence relation if needed. Would you like to explore further steps or perhaps analyze more properties of this recurrence?

Relative Questions:

1. What would the next term $a_4$ be using this recurrence relation?
2. Can you express a general formula for $a_n$?
3. How does the recurrence relation behave for large $n$?
4. What would change if the initial conditions were different?
5. Can this recurrence relation be solved using generating functions?

Tip:

When dealing with recurrence relations, solving for the first few terms helps identify patterns that might guide you towards a general solution.