1. **Problem:** Find the first 4 terms in the power series expansion of $$f(x) = \frac{(1+2x)^2}{(1-x)^2}$$ and state when the expansion is valid. 2. **Formula and rules:** - Use binomial series expansion for terms like $(1-x)^{-2}$ which is $$\sum_{n=0}^\infty \binom{n+1}{1} x^n = 1 + 2x + 3x^2 + 4x^3 + \cdots$$ - Expand numerator $(1+2x)^2 = 1 + 4x + 4x^2$ - Multiply numerator expansion by denominator expansion. - The binomial series for $(1-x)^{-2}$ converges for $|x| < 1$. 3. **Intermediate work:** - Numerator: $1 + 4x + 4x^2$ - Denominator expansion: $1 + 2x + 3x^2 + 4x^3 + \cdots$ - Multiply terms up to $x^3$: $$f(x) = (1 + 4x + 4x^2)(1 + 2x + 3x^2 + 4x^3)$$ - Calculate each term: - Constant: $1 \times 1 = 1$ - $x$: $1 \times 2x + 4x \times 1 = 2x + 4x = 6x$ - $x^2$: $1 \times 3x^2 + 4x \times 2x + 4x^2 \times 1 = 3x^2 + 8x^2 + 4x^2 = 15x^2$ - $x^3$: $1 \times 4x^3 + 4x \times 3x^2 + 4x^2 \times 2x = 4x^3 + 12x^3 + 8x^3 = 24x^3$ 4. **Result:** $$f(x) = 1 + 6x + 15x^2 + 24x^3 + \cdots$$ 5. **Validity:** The expansion is valid for $|x| < 1$ because the binomial series for $(1-x)^{-2}$ converges in this interval.