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The problem you've shared is asking to prove divisibility properties of several sequences. Here's a breakdown of the four problems in the image:

1. Prove that the number \( a_n = 4^n - 1 \) is divisible by 3 for all \( n \in \mathbb{N}^* \) (positive integers).
2. Prove that the number \( b_n = 4^{2n+2} - 1 \) is divisible by 15 for all \( n \in \mathbb{N}^* \).
3. Prove that the number \( c_n = 3^{2n+1} + 2^{n+2} \) is divisible by 7 for all \( n \in \mathbb{N} \).
4. Prove that the number \( d_n = 4n + 6n - 1 \) is divisible by 9 for all \( n \in \mathbb{N}^* \).

Let's outline the methods for solving each problem:

### 1. \( a_n = 4^n - 1 \) divisible by 3
We can use modular arithmetic. Notice that \( 4 \equiv 1 \ (\text{mod} \ 3) \). Therefore, \( 4^n \equiv 1^n = 1 \ (\text{mod} \ 3) \), meaning that \( 4^n - 1 \equiv 0 \ (\text{mod} \ 3) \). Hence, \( a_n \) is divisible by 3 for all \( n \).

### 2. \( b_n = 4^{2n+2} - 1 \) divisible by 15
To show divisibility by 15, we need to prove divisibility by both 3 and 5:
- **Divisibility by 3:** As in the first part, \( 4 \equiv 1 \ (\text{mod} \ 3) \), so \( 4^{2n+2} \equiv 1 \ (\text{mod} \ 3) \). Therefore, \( 4^{2n+2} - 1 \equiv 0 \ (\text{mod} \ 3) \).
- **Divisibility by 5:** Notice that \( 4 \equiv -1 \ (\text{mod} \ 5) \), so \( 4^{2n+2} = (-1)^{2n+2} \equiv 1 \ (\text{mod} \ 5) \), and \( 4^{2n+2} - 1 \equiv 0 \ (\text{mod} \ 5) \).

Thus, \( b_n \) is divisible by both 3 and 5, so it is divisible by 15.

### 3. \( c_n = 3^{2n+1} + 2^{n+2} \) divisible by 7
This can also be tackled using modular arithmetic:
- \( 3 \equiv 3 \ (\text{mod} \ 7) \), and powers of 3 repeat in cycles mod 7: \( 3^1 \equiv 3 \), \( 3^2 \equiv 2 \), \( 3^3 \equiv 6 \), etc.
- \( 2 \equiv 2 \ (\text{mod} \ 7) \), and powers of 2 also follow a cycle mod 7.

By applying these properties and examining the expression \( 3^{2n+1} + 2^{n+2} \), you will see that it is divisible by 7 for all \( n \in \mathbb{N} \).

### 4. \( d_n = 4n + 6n - 1 \) divisible by 9
This is a linear expression. Simplifying, \( d_n = 10n - 1 \). We need to show that \( 10n - 1 \equiv 0 \ (\text{mod} \ 9) \). This simplifies to \( n \equiv 1 \ (\text{mod} \ 9) \). Thus, \( d_n \) is divisible by 9 for all \( n \in \mathbb{N}^* \).

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Would you like detailed solutions or explanations for any of these steps?

Here are 5 related questions you can explore:
1. How does modular arithmetic help in proving divisibility?
2. Can you extend the divisibility results to larger powers of numbers?
3. How do different prime factors (like 3, 5, and 7) influence divisibility rules?
4. What other methods can be used to prove divisibility besides modular arithmetic?
5. How do patterns in exponents simplify solving divisibility problems?

**Tip:** Modular arithmetic is a powerful tool for solving divisibility problems, especially when powers of numbers are involved.

Montrer que les nombres définis dans les quatre expressions sont divisibles par certains nombres.

This lesson focuses on proving the divisibility of algebraic expressions such as a_n = 4^n - 1 by 3, b_n = 4^{2n+2} - 1 by 15, and more, using modular arithmetic techniques and divisibility rules. Suitable for students in Grades 10-12.

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