The coefficient of x3 in the expansion of 1-2x2x+12-x

1-2x2x+12-x=A2x+1+B2-x 1-2x=A(2-x)+B(2x+1) x=-12⇒2=A52⇒A=45 x=2⇒-3=B(5)⇒B=-35 ⇒45(1+2x)-1-35(2-x)-1 ⇒45(1+2x)-1-3101-x2-1 ⇒451-2x+(2x)2-(2x)3+…-3101+x2+x22+x23+… co eff of x3=45(-8)-31018 =-325-380=-51580 =-10316

The coefficient of x3 in the expansion of 1-2x2x+12-x

1-2x2x+12-x=A2x+1+B2-x 1-2x=A(2-x)+B(2x+1) x=-12⇒2=A52⇒A=45 x=2⇒-3=B(5)⇒B=-35 ⇒45(1+2x)-1-35(2-x)-1 ⇒45(1+2x)-1-3101-x2-1 ⇒451-2x+(2x)2-(2x)3+…-3101+x2+x22+x23+… co eff of x3=45(-8)-31018 =-325-380=-51580 =-10316

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The coefficient of x³ in the expansion of $\frac{1 - 2 x}{(2 x + 1) (2 - x)}$

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$- \frac{103}{16}$

$\frac{509}{80}$

$- \frac{509}{80}$

$\frac{103}{16}$

answer is C.

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Detailed Solution

$\frac{1 - 2 x}{(2 x + 1) (2 - x)} = \frac{A}{2 x + 1} + \frac{B}{2 - x} 1 - 2 x = A (2 - x) + B (2 x + 1) x = \frac{- 1}{2} \Rightarrow 2 = A (\frac{5}{2}) \Rightarrow A = \frac{4}{5} x = 2 \Rightarrow - 3 = B (5) \Rightarrow B = \frac{- 3}{5} \Rightarrow \frac{4}{5} {(1 + 2 x)}^{- 1} - \frac{3}{5} {(2 - x)}^{- 1} \Rightarrow \frac{4}{5} {(1 + 2 x)}^{- 1} - \frac{3}{10} {(1 - \frac{x}{2})}^{- 1} \Rightarrow \frac{4}{5} (1 - 2 x + {(2 x)}^{2} - {(2 x)}^{3} + \dots - \frac{3}{10} (1 + \frac{x}{2} + {(\frac{x}{2})}^{2} + {(\frac{x}{2})}^{3} + \dots)) co eff of x^{3} = \frac{4}{5} (- 8) - \frac{3}{10} (\frac{1}{8}) = \frac{- 32}{5} - \frac{3}{80} = \frac{- 515}{80} = - \frac{103}{16} \begin{array}{rcl} \end{array}$