Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
Use the binomial series to expand the function as a power series.

\[ f(x) = \frac{1}{(2-x)^2} \]

Options:

1. \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{2} \binom{-2}{n} \left(-\frac{x}{4}\right)^n\)

2. \(\displaystyle \sum_{n=0}^{\infty} \binom{-2}{n} \left(-\frac{x}{4}\right)^n\)

3. \(\displaystyle \sum_{n=0}^{\infty} \binom{-2}{n} \left(-\frac{x}{2}\right)^n\) (Selected)

4. \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{4} \binom{-2}{n} \left(-\frac{x}{2}\right)^n\) (Incorrect - indicated by a red cross)

Explanation: The problem requires expanding the given function \( f(x) = \frac{1}{(2-x)^2} \) using the binomial series. The correct expression for the power series is option 3, where the series is expressed in terms of \((-2)\), \(n\), \((-x/2)\), and the binomial coefficient.
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Transcribed Image Text:Use the binomial series to expand the function as a power series. \[ f(x) = \frac{1}{(2-x)^2} \] Options: 1. \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{2} \binom{-2}{n} \left(-\frac{x}{4}\right)^n\) 2. \(\displaystyle \sum_{n=0}^{\infty} \binom{-2}{n} \left(-\frac{x}{4}\right)^n\) 3. \(\displaystyle \sum_{n=0}^{\infty} \binom{-2}{n} \left(-\frac{x}{2}\right)^n\) (Selected) 4. \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{4} \binom{-2}{n} \left(-\frac{x}{2}\right)^n\) (Incorrect - indicated by a red cross) Explanation: The problem requires expanding the given function \( f(x) = \frac{1}{(2-x)^2} \) using the binomial series. The correct expression for the power series is option 3, where the series is expressed in terms of \((-2)\), \(n\), \((-x/2)\), and the binomial coefficient.
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