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The image contains a math problem written in Arabic and covers topics related to recursive sequences. Below is a breakdown of the key information:

1. **Recurrence Relation**: 
   - The sequence \( U_n \) is defined recursively as:
     \[
     U_{n+1} = \frac{3U_n}{U_n - 2}
     \]
     with the initial condition \( U_0 = -1 \).

2. **First Part**:
   - Calculate the first few terms of the sequence: \( U_1, U_2 \).

3. **Second Part**:
   - Another sequence \( V_n \) is introduced, where \( V_n = U_n - 1 \). 
   - The task involves defining the recursive relationship for \( V_n \) based on \( U_n \), finding the first term, and expressing the general term for \( V_n \) in terms of \( n \).

4. **Third Part**:
   - Prove that \( U_n \) is related to \( (-4) \times 3^n \), and the relation between \( U_{n+1} - U_n \).

5. **Fourth Part**:
   - Sum the sequence terms from \( U_0 \) to \( U_n \) and express the sum in terms of \( n \).

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If you'd like, I can proceed to solve these parts. Would you like a detailed solution for each part? Here are a few relevant questions to consider:

1. How do we compute \( U_1 \) and \( U_2 \) using the given recurrence relation?
2. What is the recursive relation for \( V_n \) in terms of \( U_n \)?
3. How does the sequence \( V_n \) help in simplifying the analysis of \( U_n \)?
4. How is the sequence sum \( U_0 + U_1 + \cdots + U_n \) related to \( n \)?
5. Can we derive the general term for the sequence using the given relation between \( U_n \) and powers of 3?

**Tip**: When solving recursive sequences, calculating the first few terms is often key to identifying a pattern or simplifying the problem.

Recurrence sequence defined by U_(n+1) = (3U_n) / (U_n - 2), with U_0 = -1. Calculate U_1, U_2, define a related sequence V_n = U_n - 1, and analyze the convergence.

This problem involves a recurrence sequence where U_(n+1) is defined by (3U_n) / (U_n - 2), starting with U_0 = -1. Explore how to calculate U_1, U_2, define the related sequence V_n, and analyze convergence. Suitable for advanced high school students.

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