1. **Problem statement:** (b) Express $ (2 + x)^{-1} $ in the form $ A(1 + Bx)^{-1} $ where $ A $ and $ B $ are rational numbers. (c) Given $ f(x) = \frac{1 + 4x}{(2 + x)^3} $, obtain a series expansion for $ f(x) $ in ascending powers of $ x $ up to and including the term in $ x^2 $. --- 2. **Part (b) solution:** We want to write: $$ (2 + x)^{-1} = A(1 + Bx)^{-1} $$ Rewrite $ 2 + x $ as $ 2(1 + \frac{x}{2}) $: $$ (2 + x)^{-1} = \frac{1}{2(1 + \frac{x}{2})} = \frac{1}{2} (1 + \frac{x}{2})^{-1} $$ Comparing, we get: $$ A = \frac{1}{2}, \quad B = \frac{1}{2} $$ --- 3. **Part (c) solution:** Given: $$ f(x) = \frac{1 + 4x}{(2 + x)^3} $$ Rewrite denominator using part (b) idea: $$ (2 + x)^3 = 2^3 (1 + \frac{x}{2})^3 = 8 (1 + \frac{x}{2})^3 $$ So: $$ f(x) = \frac{1 + 4x}{8 (1 + \frac{x}{2})^3} = \frac{1 + 4x}{8} (1 + \frac{x}{2})^{-3} $$ Expand $ (1 + \frac{x}{2})^{-3} $ using binomial series up to $ x^2 $: The binomial expansion for $ (1 + u)^n $ is: $$ 1 + nu + \frac{n(n-1)}{2} u^2 + \cdots $$ Here, $ n = -3 $ and $ u = \frac{x}{2} $: $$ (1 + \frac{x}{2})^{-3} = 1 - 3 \cdot \frac{x}{2} + \frac{-3(-4)}{2} \left(\frac{x}{2}\right)^2 = 1 - \frac{3x}{2} + 6 \cdot \frac{x^2}{4} = 1 - \frac{3x}{2} + \frac{3x^2}{2} $$ Multiply by $ \frac{1 + 4x}{8} $: $$ f(x) = \frac{1}{8} (1 + 4x) \left(1 - \frac{3x}{2} + \frac{3x^2}{2}\right) $$ Expand numerator: $$ (1 + 4x) \left(1 - \frac{3x}{2} + \frac{3x^2}{2}\right) = 1 \cdot \left(1 - \frac{3x}{2} + \frac{3x^2}{2}\right) + 4x \cdot \left(1 - \frac{3x}{2} + \frac{3x^2}{2}\right) $$ Calculate terms: - $ 1 \cdot 1 = 1 $ - $ 1 \cdot (-\frac{3x}{2}) = -\frac{3x}{2} $ - $ 1 \cdot \frac{3x^2}{2} = \frac{3x^2}{2} $ - $ 4x \cdot 1 = 4x $ - $ 4x \cdot (-\frac{3x}{2}) = -6x^2 $ - $ 4x \cdot \frac{3x^2}{2} = 12x^3 $ (ignore $ x^3 $ term as we only want up to $ x^2 $) Sum terms up to $ x^2 $: $$ 1 - \frac{3x}{2} + \frac{3x^2}{2} + 4x - 6x^2 = 1 + \left(-\frac{3}{2} + 4\right)x + \left(\frac{3}{2} - 6\right)x^2 = 1 + \frac{5x}{2} - \frac{9x^2}{2} $$ Multiply by $ \frac{1}{8} $: $$ f(x) = \frac{1}{8} + \frac{5x}{16} - \frac{9x^2}{16} $$ --- **Final answers:** (b) $ A = \frac{1}{2} $, $ B = \frac{1}{2} $ so $$ (2 + x)^{-1} = \frac{1}{2} (1 + \frac{x}{2})^{-1} $$ (c) Series expansion of $ f(x) $ up to $ x^2 $: $$ f(x) = \frac{1}{8} + \frac{5}{16} x - \frac{9}{16} x^2 $$