Question

The coefficient of \(x^2\) in the power series expansion of \[ \frac{2x}{(x-1)(x+2)^2} \] when \(|x|<1\) is 

• A. \( 0 \)
• B. \( -\frac{2}{3} \)
• C. \( \frac{4}{9} \)
• D. \( \frac{1}{3} \)

Hint:

For \( (1+kx)^{-n} \), the coefficient of \( x^r \) is \( (-1)^r \binom{n+r-1}{r} k^r \). Using this general formula helps avoid manual expansion errors.

Answer 1

The Correct Option is A

Concept: To find the coefficient in a power series, decompose the function using partial fractions and expand each term using the geometric series formula:
• \( (1-x)^{-1} = 1 + x + x^2 + \dots \)
• \( (1+x)^{-2} = 1 - 2x + 3x^2 - \dots \)

Step 1: Decomposing into partial fractions.
Let \( \frac{2x}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} \). Multiplying by the denominator: \( 2x = A(x+2)^2 + B(x-1)(x+2) + C(x-1) \).
• Put \( x = 1 \): \( 2 = 9A \implies A = 2/9 \).
• Put \( x = -2 \): \( -4 = -3C \implies C = 4/3 \).
• Compare coeff. of \( x^2 \): \( 0 = A + B \implies B = -2/9 \).

Step 2: Expanding the terms.
1) \( \frac{2/9}{x-1} = -\frac{2}{9}(1-x)^{-1} = -\frac{2}{9}(1+x+x^2+\dots) \). Coeff of \( x^2 = -2/9 \). 2) \( \frac{-2/9}{x+2} = -\frac{2}{9} \cdot \frac{1}{2(1+x/2)} = -\frac{1}{9}(1 - \frac{x}{2} + \frac{x^2}{4} - \dots) \). Coeff of \( x^2 = -1/36 \). 3) \( \frac{4/3}{(x+2)^2} = \frac{4}{3} \cdot \frac{1}{4(1+x/2)^2} = \frac{1}{3}(1 - 2(\frac{x}{2}) + 3(\frac{x^2}{4}) - \dots) \). Coeff of \( x^2 = 1/4 \).

Step 3: Summing the coefficients.
Total coefficient of \( x^2 = -\frac{2}{9} - \frac{1}{36} + \frac{1}{4} = \frac{-8 - 1 + 9}{36} = 0 \).