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The problem presents a sequence defined by \( u_0 = 12 \) and the recurrence relation:

\[
u_{n+1} = \frac{1}{3}u_n + 2
\]

### Question 1: Using the graph, conjecture the behavior of the sequence \((u_n)\).
The graph shows two lines:
1. **Blue line**: \( y = x \), which is the identity line. It can represent a fixed value where \( u_{n+1} = u_n \).
2. **Red line**: \( y = \frac{1}{3}x + 2 \), which represents the recurrence relation.

Observing the graph, you can see that the sequence starts at \( u_0 = 12 \) and the next values of \( u_n \) are generated by applying the recurrence relation. Since the slope of the line \( y = \frac{1}{3}x + 2 \) is less than 1, the sequence will decrease over time but approach a limiting value (asymptote).

#### Conjecture:
The sequence \((u_n)\) is **decreasing** and will eventually stabilize towards a limit, which can be guessed from the intersection of the lines. The limit is found when \( u_{n+1} = u_n \), so:

\[
u = \frac{1}{3}u + 2
\]

Solving for \( u \):

\[
u - \frac{1}{3}u = 2 \implies \frac{2}{3}u = 2 \implies u = 3
\]

Thus, the sequence tends towards the value **3**.

### Question 2: Prove the conjecture by induction.

We will prove that the sequence is decreasing, i.e., \( u_{n+1} \leq u_n \), and show that it converges to the value 3.

#### Base Case:
For \( n = 0 \):

\[
u_1 = \frac{1}{3}u_0 + 2 = \frac{1}{3} \times 12 + 2 = 4 + 2 = 6
\]

Clearly, \( u_1 = 6 \leq 12 = u_0 \).

#### Induction Hypothesis:
Assume that \( u_n \leq u_{n-1} \) for some \( n \geq 1 \).

#### Induction Step:
We need to show that \( u_{n+1} \leq u_n \). From the recurrence relation:

\[
u_{n+1} = \frac{1}{3}u_n + 2
\]

We know that \( u_n \geq 3 \), since the sequence is decreasing and approaches 3. We need to check that \( u_{n+1} \leq u_n \):

\[
u_{n+1} = \frac{1}{3}u_n + 2 \leq u_n
\]

This simplifies to:

\[
\frac{1}{3}u_n + 2 \leq u_n
\]

Subtracting \( \frac{1}{3}u_n \) from both sides:

\[
2 \leq \frac{2}{3}u_n
\]

Multiplying both sides by \( \frac{3}{2} \):

\[
3 \leq u_n
\]

Thus, the sequence is decreasing and bounded below by 3. Therefore, by the principle of mathematical induction, the sequence converges to 3.

---

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. What happens to the sequence if the initial value \( u_0 \) is changed?
2. How would the recurrence relation change if the constant 2 in \( u_{n+1} = \frac{1}{3}u_n + 2 \) were different?
3. How can you determine if a recurrence relation has a limit without graphing it?
4. Can you express the general term \( u_n \) explicitly without using recurrence?
5. What other types of sequences have similar long-term behavior (e.g., geometric or arithmetic)?

**Tip:** When working with recurrence relations, try solving for the fixed point (limit) by setting \( u_{n+1} = u_n \).

We are given a sequence defined by u_0 = 12 and for all n in natural numbers, u_{n+1} = (1/3)u_n + 2. Using the graph, conjecture the behavior of the sequence (u_n). Prove the conjecture using mathematical induction.

This problem focuses on a sequence defined by a recurrence relation u_{n+1} = (1/3)u_n + 2 with an initial value u_0 = 12. Explore how to conjecture the behavior of the sequence using graphs and prove the conjecture using mathematical induction. Suitable for high school students in Grades 10-12.

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