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Let's go through the problem step by step to solve it.

### Exercise 4 Overview:
We are given a sequence \( (u_n) \) defined by:
- \( u_0 = 0 \)
- \( u_{n+1} = \frac{2}{5} u_n + 1 \), for \( n \in \mathbb{N} \)

The tasks are as follows:

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### 1. Show by induction that \( u_n \leq \frac{5}{3} \) for all \( n \in \mathbb{N} \).

#### Induction Proof Outline:
- **Base case:** Show that \( u_0 \leq \frac{5}{3} \).
- **Inductive hypothesis:** Assume \( u_n \leq \frac{5}{3} \) holds for some \( n = k \).
- **Inductive step:** Show that \( u_{n+1} = \frac{2}{5} u_n + 1 \leq \frac{5}{3} \) under the assumption.

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### 2. a) Show that \( u_{n+1} - u_n = -\frac{3}{5} \left( u_n - \frac{5}{3} \right) \).

We need to manipulate the recurrence relation \( u_{n+1} = \frac{2}{5} u_n + 1 \) to express the difference \( u_{n+1} - u_n \) in terms of \( u_n \).

b) Deduce that the sequence \( (u_n) \) is increasing.

Using the result from part (a), we analyze the sign of \( u_{n+1} - u_n \) to determine whether the sequence is increasing.

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### 3. Define \( v_n = u_n - \frac{5}{3} \) for all \( n \in \mathbb{N} \).

This creates a new sequence \( (v_n) \), and the following tasks are based on this definition:

a) Calculate \( v_0 \).

b) Show that the sequence \( (v_n) \) is geometric with a ratio \( q = \frac{2}{5} \).

c) Express \( v_n \) as a function of \( n \), and deduce the general formula for \( u_n \).

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### Let's start solving each step systematically.

Would you like a detailed solution for each part?

Considérons la suite (u_n) définie par: u_0 = 0 et u_(n+1) = (2/5)u_n + 1 pour tout n de N. Montrer que la suite (u_n) est croissante et déterminer son comportement asymptotique.

Explore how to solve a recursive sequence u_n with the recurrence relation u_(n+1) = (2/5)u_n + 1. This solution covers key concepts such as geometric sequences, proving monotonicity, and finding the asymptotic behavior of the sequence. Suitable for advanced high school students.

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