$\begin{align} & \text{Express the function }f(x)=\frac{11{{x}^{2}}-12x+4}{{{(x-1)}^{2}}(4x-1)}\text{ in partial fractions}\text{. } \\ & \text{Hence or otherwise expand }f(x)\text{ in the form} \\ & f(x)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+\cdot \cdot \cdot +{{a}_{n}}{{x}^{n}}\cdot \cdot \cdot \\ & \text{State the value of }x\text{ for which the expansion is valid} \\\end{align}$
$\begin{align} & \frac{11{{x}^{2}}-12x+4}{{{(x-1)}^{2}}(4x-1)}=\frac{A}{x-1}+\frac{B}{{{(x-1)}^{2}}}+\frac{C}{(4x-1)} \\ & A(x-1)(4x-1)+B(4x-1)+C{{(x-1)}^{2}}=11{{x}^{2}}-12x+4--(i) \\ & \text{Set }x=1 \\ & 3B=11-12+4;\text{ }B=1 \\ & \text{Set }x=\frac{1}{4} \\ & C{{\left( \frac{1}{4}-1 \right)}^{2}}=11{{\left( \frac{1}{4} \right)}^{2}}-12\left( \frac{1}{4} \right)+4 \\ & \frac{9}{16}C=\frac{11}{16}-3+4 \\ & \frac{9}{16}C=\frac{11+16}{16} \\ & \frac{9}{16}C=\frac{27}{16} \\ & C=3 \\ & \text{Set }x=0\text{ in the equation, substitute }B=1,\text{ and }C=3 \\ & A-B+C=4 \\ & A-1+3=4 \\ & A=2 \\ & f(x)=\frac{2}{x-1}+\frac{1}{{{(x-1)}^{2}}}+\frac{3}{(4x-1)} \\ & f(x)=2{{(x-1)}^{-1}}+{{(x-1)}^{-2}}+3{{(4x-1)}^{-1}} \\ & f(x)=2{{(x-1)}^{-1}}+{{(x-1)}^{-2}}+3{{[-1(1-4x)]}^{-1}} \\ & f(x)=2{{(x-1)}^{-1}}+{{(x-1)}^{-2}}+3[-{{1}^{-1}}{{(1-4x)}^{-1}}] \\ & f(x)=2{{(x-1)}^{-1}}+{{(x-1)}^{-2}}-3{{(1-4x)}^{-1}} \\ & \text{Expanding this will give } \\ & f(x)=-4-12x-47{{x}^{2}}+----- \\ & \text{The expansion is valid when }\left| x \right|<\tfrac{1}{4} \\\end{align}$