Question

The coefficient of \(x^3\) in the power series expansion of \[ \frac{x}{x^2-x-2} \] is

• A. \(-\dfrac{1}{8}\)
• B. \(-\dfrac{3}{8}\)
• C. \(\dfrac{3}{8}\)
• D. \(\dfrac{1}{8}\)

Hint:

For power series expansions of rational functions, first use partial fractions. Then convert each fraction into a geometric series and extract the required coefficient.

Answer 1

The Correct Option is B

Concept: To find the coefficient of a particular power of \(x\), first decompose the rational function into partial fractions and then use the geometric series expansion \[ \frac{1}{1-r}=1+r+r^2+r^3+\cdots \] for \(|r|\lt 1\).

Step 1: Decompose into partial fractions. \[ \frac{x}{x^2-x-2} = \frac{x}{(x-2)(x+1)}. \] Let \[ \frac{x}{(x-2)(x+1)} = \frac{A}{x-2}+\frac{B}{x+1}. \] Then \[ x=A(x+1)+B(x-2). \] Comparing coefficients, \[ A+B=1, \] \[ A-2B=0. \] Solving, \[ A=\frac23, \qquad B=\frac13. \] Hence, \[ \frac{x}{x^2-x-2} = \frac{2}{3(x-2)} + \frac{1}{3(x+1)}. \]

Step 2: Expand each term as a power series. \[ \frac{2}{3(x-2)} = -\frac13\cdot\frac{1}{1-\frac{x}{2}}. \] Using \[ \frac{1}{1-\frac{x}{2}} = 1+\frac{x}{2}+\frac{x^2}{2^2}+\frac{x^3}{2^3}+\cdots, \] we get \[ \frac{2}{3(x-2)} = -\frac13 -\frac{x}{6} -\frac{x^2}{12} -\frac{x^3}{24} +\cdots. \] Also, \[ \frac{1}{3(x+1)} = \frac13\cdot\frac{1}{1+x}. \] Using \[ \frac{1}{1+x} = 1-x+x^2-x^3+\cdots, \] we get \[ \frac{1}{3(x+1)} = \frac13 -\frac{x}{3} +\frac{x^2}{3} -\frac{x^3}{3} +\cdots. \]

Step 3: Find the coefficient of \(x^3\). Coefficient of \(x^3\) from the first series: \[ -\frac{1}{24}. \] Coefficient of \(x^3\) from the second series: \[ -\frac13. \] Therefore, \[ -\frac{1}{24}-\frac13 = -\frac{1}{24}-\frac{8}{24} = -\frac{9}{24} = -\frac38. \] Hence, the coefficient of \(x^3\) is \[ \boxed{-\frac38} \] \[ \boxed{\text{Answer = (B)}} \]