$\begin{align}  & \frac{1}{(x+2)(2x+1)}=\frac{A}{(x+2)}+\frac{B}{(2x+1)} \\ & 1=A(2x+1)+B(x+2) \\ & \text{Let }x=-2 \\ & 1=5A,\text{   }A=-\frac{1}{3} \\ & \text{Put }x=-\frac{1}{2} \\ & 1=B(-\tfrac{1}{2}+2) \\ & \frac{3}{2}B=1 \\ & B=\frac{2}{3} \\ & \text{The partial fraction becomes} \\ & \frac{1}{(x+2)(2x+1)}=-\frac{1}{3(x+2)}+\frac{2}{3(2x+1)} \\ & \frac{1}{(x+2)(2x+1)}=\frac{1}{3}\left( \frac{2}{(2x+1)}-\frac{1}{x+2} \right) \\ & \frac{1}{3}\left( \frac{2}{(2x+1)}-\frac{1}{x+2} \right)=\frac{1}{3}\left( 2{{(1+2x)}^{-1}}-{{(2+x)}^{-1}} \right) \\ & \frac{1}{3}\left( \frac{2}{(2x+1)}-\frac{1}{x+2} \right)=\frac{1}{3}\left[ 2{{(1+2x)}^{-1}}-\left( {{2}^{-1}}{{(1+\tfrac{x}{2})}^{-1}} \right) \right] \\ & \frac{1}{3}\left( \frac{2}{(2x+1)}-\frac{1}{x+2} \right)=\frac{2}{3}{{(1+2x)}^{-1}}-\frac{1}{6}{{(1+\tfrac{x}{2})}^{-1}} \\ & =\frac{2}{3}\left( 1-2x+4{{x}^{2}}-8{{x}^{3}}+... \right)-\frac{1}{6}\left( 1-\frac{x}{2}+\frac{{{x}^{2}}}{4}-\frac{{{x}^{3}}}{8}+... \right) \\ & \text{=}\frac{2}{3}-\frac{4}{3}x+\frac{8}{3}{{x}^{2}}-\frac{16}{3}{{x}^{3}}-\frac{1}{6}+\frac{1}{12}x-\frac{1}{24}{{x}^{2}}+\frac{1}{48}{{x}^{3}}+... \\ & =\frac{1}{2}-\frac{5}{4}x+\frac{21}{8}{{x}^{2}}+\frac{1}{48}{{x}^{3}}+----{{(-1)}^{n}}\left[ \sum\limits_{r=1}^{n}{\frac{{{2}^{2n-2}}}{{{2}^{n}}}} \right]{{x}^{n-1}} \\ & \text{The coefficient of }{{x}^{n}}\text{ is }{{(-1)}^{n}}\left[ \sum\limits_{r=1}^{n}{\frac{{{2}^{2n-2}}}{{{2}^{n}}}} \right]\frac{1}{x} \\\end{align}$