The equation provided in the image is a recurrence relation:

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

with the initial conditions:

$a_1 = 1, \quad a_2 = -1, \quad a_3 = 0.$

This is a linear recurrence relation, and we can calculate further terms by using the given formula.

Let's calculate the next few terms $a_4, a_5, a_6, \dots$:

For $n = 4$:

$a_4 = -a_3 + 2a_2 + 3a_1 = -(0) + 2(-1) + 3(1) = -2 + 3 = 1$

For $n = 5$:

$a_5 = -a_4 + 2a_3 + 3a_2 = -(1) + 2(0) + 3(-1) = -1 + 0 - 3 = -4$

For $n = 6$:

$a_6 = -a_5 + 2a_4 + 3a_3 = -(-4) + 2(1) + 3(0) = 4 + 2 = 6$

The sequence starts: $a_1 = 1, a_2 = -1, a_3 = 0, a_4 = 1, a_5 = -4, a_6 = 6$.

Would you like more terms or any further clarification?

Related Questions:

1. What methods can solve recurrence relations like this one?
2. How would the solution differ for non-homogeneous recurrence relations?
3. Can we express this sequence with a closed-form solution?
4. What applications are there for recurrence relations in computer science or mathematics?
5. How does this recurrence relate to Fibonacci or other famous sequences?

Tip:

Recurrence relations can often be solved using characteristic equations if the relation is linear with constant coefficients.