Question

When \( |x| < 2 \), the coefficient of \( x^2 \) in the power series expansion of

\[ \frac{x}{(x-2)(x-3)} \]

is:

• A. \(\frac{1}{6}\)
• B. \(\frac{5}{36}\)
• C. \(\frac{25}{216}\)
• D. \(\frac{5}{18}\)

Hint:

Use partial fractions to simplify complex rational functions before expanding them into power series, especially for finding specific coefficients.

Answer 1

The Correct Option is B

We can use partial fractions to decompose the expression: \[\frac{x}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}\] \[x = A(x-3) + B(x-2)\] Let \(x = 2\): \[2 = A(2-3) + B(2-2)\] \[2 = -A \implies A = -2\] Let \(x = 3\): \[3 = A(3-3) + B(3-2)\] \[3 = B\] So, \[\frac{x}{(x-2)(x-3)} = \frac{-2}{x-2} + \frac{3}{x-3}\] \[= \frac{2}{2-x} - \frac{3}{3-x}\] \[= \frac{2}{2(1-\frac{x}{2})} - \frac{3}{3(1-\frac{x}{3})}\] \[= \frac{1}{1-\frac{x}{2}} - \frac{1}{1-\frac{x}{3}}\] Since \(|x|<2\), we have \(\left|\frac{x}{2}\right|<1\) and \(\left|\frac{x}{3}\right|<1\). We can use the geometric series expansion: \[\frac{1}{1-r} = 1 + r + r^2 + r^3 + \dots\] Thus, \[\frac{1}{1-\frac{x}{2}} = 1 + \frac{x}{2} + \left(\frac{x}{2}\right)^2 + \left(\frac{x}{2}\right)^3 + \dots\] \[\frac{1}{1-\frac{x}{3}} = 1 + \frac{x}{3} + \left(\frac{x}{3}\right)^2 + \left(\frac{x}{3}\right)^3 + \dots\] So, \[\frac{x}{(x-2)(x-3)} = \left(1 + \frac{x}{2} + \frac{x^2}{4} + \dots\right) - \left(1 + \frac{x}{3} + \frac{x^2}{9} + \dots\right)\] \[= \left(\frac{1}{2} - \frac{1}{3}\right)x + \left(\frac{1}{4} - \frac{1}{9}\right)x^2 + \dots\] \[= \left(\frac{3-2}{6}\right)x + \left(\frac{9-4}{36}\right)x^2 + \dots\] \[= \frac{1}{6}x + \frac{5}{36}x^2 + \dots\] The coefficient of \(x^2\) is \(\frac{5}{36}\). Final Answer: The final answer is $\boxed{(2)}$