https://file.aiquickdraw.com/mathbot/file/1728051253136or37nk3t.jpg

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.

Given the recurrence relation a_n = -a_(n-1) + 2a_(n-2) + 3a_(n-3) with initial conditions a_1 = 1, a_2 = -1, and a_3 = 0, calculate the next few terms in the sequence.

Explore how to solve the recurrence relation a_n = -a_(n-1) + 2a_(n-2) + 3a_(n-3) with initial conditions a_1 = 1, a_2 = -1, and a_3 = 0. Step-by-step solutions to find the next few terms of the sequence. Suitable for discrete mathematics students.

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