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The problem in *Exercice 1* involves proving by induction that for all integers \( n \geq 2 \), the term \( u_n \) in the sequence defined by:

\[
u_2 = 3 \quad \text{and} \quad u_{n+1} = \frac{3u_n + 1}{u_n + 3}, \quad \text{for } n \geq 2
\]

can be expressed as:

\[
u_n = \frac{2^n + 2}{2^n - 2}.
\]

### Plan for Inductive Proof:

1. **Base Case**: Verify that the formula holds for \( n = 2 \).

2. **Inductive Hypothesis**: Assume the formula holds for \( n = k \), i.e.,

   \[
   u_k = \frac{2^k + 2}{2^k - 2}.
   \]

3. **Inductive Step**: Show that the formula holds for \( n = k + 1 \), i.e.,

   \[
   u_{k+1} = \frac{2^{k+1} + 2}{2^{k+1} - 2}.
   \]

We begin by verifying the base case and then proceed with the inductive steps.

Would you like a detailed step-by-step breakdown for each of these parts?

Démontrer par récurrence que pour tout entier n ≥ 2 on a un = (2^n + 2)/(2^n - 2), avec u2 = 3 et un+1 = (3un + 1)/(un + 3) pour tout n ≥ 2.

Learn how to prove by induction that the sequence defined by u2 = 3 and un+1 = (3un + 1)/(un + 3) satisfies the formula un = (2^n + 2)/(2^n - 2) for all n ≥ 2. A step-by-step breakdown is provided for advanced high school or college students.

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