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?

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 $x$.