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We are asked to expand \( \frac{(1 + 2x)^2}{(2 - x)} \) in ascending powers of \(x\) up to and including the term in \(x^3\), and to state the values of \(x\) for which the expansion is valid.

### Step-by-Step Solution:

1. **First, expand the numerator** \((1 + 2x)^2\):
   \[
   (1 + 2x)^2 = 1^2 + 2(1)(2x) + (2x)^2 = 1 + 4x + 4x^2
   \]

2. **Next, express the denominator** \(\frac{1}{2 - x}\) as a binomial series expansion. We want to expand this in powers of \(x\), treating it like \( \frac{1}{a - x} \), where \( a = 2 \). Using the binomial expansion for \( \frac{1}{a - x} = \frac{1}{a} \left(1 - \frac{x}{a}\right)^{-1} \), we have:
   \[
   \frac{1}{2 - x} = \frac{1}{2} \left( 1 - \frac{x}{2} \right)^{-1}
   \]
   Using the binomial series for \( (1 - z)^{-1} = 1 + z + z^2 + z^3 + \cdots \) when \( |z| < 1 \), we get:
   \[
   \frac{1}{2 - x} = \frac{1}{2} \left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \cdots \right)
   \]
   Simplifying:
   \[
   \frac{1}{2 - x} = \frac{1}{2} + \frac{x}{4} + \frac{x^2}{8} + \frac{x^3}{16} + \cdots
   \]

3. **Multiply the expanded numerator and denominator**:
   We now multiply \((1 + 4x + 4x^2)\) by the expansion of \(\frac{1}{2 - x}\):
   \[
   (1 + 4x + 4x^2)\left( \frac{1}{2} + \frac{x}{4} + \frac{x^2}{8} + \frac{x^3}{16} + \cdots \right)
   \]

   Distribute each term:
   - \( 1 \times \left(\frac{1}{2} + \frac{x}{4} + \frac{x^2}{8} + \frac{x^3}{16}\right) = \frac{1}{2} + \frac{x}{4} + \frac{x^2}{8} + \frac{x^3}{16} \)
   - \( 4x \times \left(\frac{1}{2} + \frac{x}{4} + \frac{x^2}{8}\right) = 2x + x^2 + \frac{x^3}{2} \)
   - \( 4x^2 \times \left(\frac{1}{2} + \frac{x}{4}\right) = 2x^2 + x^3 \)

4. **Add the terms**:
   Now combine all like terms up to \(x^3\):
   \[
   \frac{1}{2} + \frac{x}{4} + \frac{x^2}{8} + \frac{x^3}{16} + 2x + x^2 + \frac{x^3}{2} + 2x^2 + x^3
   \]
   Grouping the powers of \(x\):
   - Constant: \( \frac{1}{2} \)
   - Coefficient of \(x\): \( \frac{x}{4} + 2x = \frac{9x}{4} \)
   - Coefficient of \(x^2\): \( \frac{x^2}{8} + x^2 + 2x^2 = \frac{25x^2}{8} \)
   - Coefficient of \(x^3\): \( \frac{x^3}{16} + \frac{x^3}{2} + x^3 = \frac{29x^3}{16} \)

   Therefore, the expansion up to the term in \(x^3\) is:
   \[
   \frac{(1 + 2x)^2}{(2 - x)} = \frac{1}{2} + \frac{9x}{4} + \frac{25x^2}{8} + \frac{29x^3}{16} + \cdots
   \]

5. **Radius of convergence**:
   The binomial series expansion \( \frac{1}{2 - x} \) converges when \( |x| < 2 \) (since we expanded around \(x = 0\)).

Thus, the values of \(x\) for which the expansion is valid are \( |x| < 2 \).

Would you like to explore more details on any part of the process or have any further questions?

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Here are 5 related questions to deepen understanding:
1. How would the expansion change if the numerator were \( (1 - 2x)^2 \) instead?
2. Can you use this method to expand up to \(x^4\)?
3. How does the general binomial expansion formula for \( (1 - z)^{-n} \) work for non-integer values of \(n\)?
4. What happens if you expand around a different center, say \(x = 1\)?
5. How would you apply this method to functions that aren't rational expressions?

**Tip**: When using binomial expansions, always check the convergence radius to ensure the expansion is valid for your input values of \(x\).

Expand (1 + 2x)^2 / (2 - x) in ascending powers of x up to and including the term in x^3 and state the values of x for which the expansion is valid.

This solution demonstrates the binomial expansion of the expression (1 + 2x)^2 / (2 - x) in ascending powers of x, including the term in x^3. It also addresses the radius of convergence, stating that the expansion is valid for |x| < 2. Ideal for advanced high school and undergraduate students.

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