Let's first look at the equation you have:
a
n
=
2
a
n
−
1
+
a
n
−
2
−
2
a
n
−
3
a 
n
​	
 =2a 
n−1
​	
 +a 
n−2
​	
 −2a 
n−3
​	
 
You mentioned some values:
a
0
=
3
a 
0
​	
 =3
a
1
=
6
a 
1
​	
 =6
a
2
=
0
a 
2
​	
 =0

Let's break down the equation you provided:

\[
a_n = 2a_{n-1} + a_{n-2} - 2a_{n-3}
\]

This is a recursive relation, which defines each term \(a_n\) in terms of the previous three terms in the sequence: \(a_{n-1}, a_{n-2},\) and \(a_{n-3}\).

You also provided initial conditions:

- \(a_0 = 3\)
- \(a_1 = 6\)
- \(a_2 = 0\)

To find further terms in the sequence (e.g., \(a_3, a_4\), etc.), we can apply the recursive relation step by step.

### Let's find \(a_3\):

Using the equation \(a_n = 2a_{n-1} + a_{n-2} - 2a_{n-3}\) for \(n = 3\):

\[
a_3 = 2a_2 + a_1 - 2a_0
\]

Substituting the known values \(a_2 = 0\), \(a_1 = 6\), and \(a_0 = 3\):

\[
a_3 = 2(0) + 6 - 2(3)
\]

\[
a_3 = 6 - 6 = 0
\]

So, \(a_3 = 0\).

### Let's find \(a_4\):

Now, for \(n = 4\):

\[
a_4 = 2a_3 + a_2 - 2a_1
\]

Substitute the known values \(a_3 = 0\), \(a_2 = 0\), and \(a_1 = 6\):

\[
a_4 = 2(0) + 0 - 2(6)
\]

\[
a_4 = 0 - 12 = -12
\]

So, \(a_4 = -12\).

### Let's find \(a_5\):

Now, for \(n = 5\):

\[
a_5 = 2a_4 + a_3 - 2a_2
\]

Substitute the known values \(a_4 = -12\), \(a_3 = 0\), and \(a_2 = 0\):

\[
a_5 = 2(-12) + 0 - 2(0)
\]

\[
a_5 = -24 + 0 = -24
\]

So, \(a_5 = -24\).

---

Would you like to see more terms of the sequence or a deeper explanation?

### Related questions:
1. How do we determine the general form of a recursive sequence?
2. Can we find an explicit formula for this recursive sequence?
3. How does changing initial conditions affect the sequence?
4. How are recursive sequences used in real-world problems?
5. What happens if we add more terms to the recursive equation?

### Tip:
When working with recursive relations, having clear initial conditions is essential to generating the correct sequence. Always verify each step carefully to avoid errors when calculating subsequent terms!

Let's first look at the equation you have: a_n = 2a_{n-1} + a_{n-2} - 2a_{n-3} You mentioned some values: a_0 = 3, a_1 = 6, a_2 = 0.

Explore the step-by-step solution of the recursive sequence a_n = 2a_{n-1} + a_{n-2} - 2a_{n-3} with initial conditions a_0 = 3, a_1 = 6, and a_2 = 0. Learn how to calculate subsequent terms and understand the underlying recurrence relation. Suitable for students in Grades 10-12.

Math Problem Statement

Let's first look at the equation you have: a n

You mentioned some values: a 0

3 a 0 =3 a 1

6 a 1 =6 a 2

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