Question

For \(|x|<\frac{1}{\sqrt{2}}\) the coefficient of \(x\) in the expansion of \(\frac{(1-4x)^2(1-2x^2)^{1/2}}{(4-x)^{3/2}}\) is

• A. \(\frac{61}{64}\)
• B. \(-\frac{61}{64}\)
• C. \(\frac{69}{64}\)
• D. \(-\frac{69}{64}\)

Hint:

Use the binomial expansion \((1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!} x^2 + \dots\)

Answer 1

The Correct Option is B

We want to find the coefficient of \(x\) in the expansion of

\[ \frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}}. \] 

We can write

\[ \frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}} = (1 - 4x)^2 (1 - 2x^2)^{1/2} \cdot 4^{-3/2} \left( 1 - \frac{x}{4} \right)^{-3/2} \] \[ = \frac{1}{8} (1 - 8x + 16x^2) (1 - x^2 + \ldots) \left( 1 + \frac{3}{8} x + \ldots \right). \]

We are only interested in the coefficient of \(x\), so we can ignore terms of degree 2 or higher. Then

\[ \frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}} = \frac{1}{8} (1 - 8x) \left( 1 + \frac{3}{8} x \right) + O(x^2) \] \[ = \frac{1}{8} \left( 1 - 8x + \frac{3}{8} x - 3x^2 \right) + O(x^2) \] \[ = \frac{1}{8} \left( 1 - \frac{61}{8} x \right) + O(x^2) \] \[ = \frac{1}{8} - \frac{61}{64} x + O(x^2). \]

Therefore, the coefficient of \(x\) is \(-\frac{61}{64}\).