<?xml version="1.0" encoding="UTF-8"?>
<quiz>
  <question type="description">
    <name>
      <text>9.2.12 (Examples of negative power binomial)</text>
    </name>
    <questiontext format="moodle_auto_format">
      <text><![CDATA[<h3>Examples</h3>

<p>Find the first three terms in the expansions of the following.</p>

<ol type="a">
<li>\(\frac{2}{(1-3x)}\) for \(|x| < \frac{1}{3}\).</li>
<li>\((2 + 3x)^{-\frac{2}{3}}\) for \(|x| < \frac{2}{3}\).</li>
</ol>

<h3>Solutions</h3>

<ol type="a">
<li>Here we write \(\frac{2}{(1-3x)} = 2(1-3x)^{-1}\). Then using the expansion formulas above we have 
\[
\begin{align*}
2(1-3x)^{-1} &= 2 \left( 1 + 3x + \frac{-1(-1-1)}{2!} (-3x^2) + \frac{-1(-1-1)(-1-2)}{3!} (-3x^3)\right)
\\
&= 2 \left( 1 + 3x + 9x^2 + 27x^3 \right)
\\
&= 2 + 6x + 18x^2 + 54x^3.
\end{align*}
\]</li>
<li>Here, we can write \((2 + 3x)^{-\frac{2}{3}} = 2^{-\frac{2}{3}}(1 + \frac{3x}{2})\). Thus we can use the formulas above with \(n = -\frac{2}{3}\) and we use \(\frac{3}{2} x\) in place of \(x\). So we can expand the binomial as follows:
\[
\begin{align*}
2^{-\frac{2}{3}}(1 + \frac{3x}{2}) &= 2^{-\frac{2}{3}} \left(1 + (-x) + \frac{\frac{2}{3}(-\frac{2}{3}-1)}{2!} (\frac{3x}{2})^2 + \frac{\frac{-2}{3}(\frac{-2}{3}-1)(\frac{-2}{3}-2)}{3!} (\frac{3x}{2})^3 \right)
\\
&= 2^{-\frac{2}{3}} \left( 1 - x + \frac{5}{4} x^2 - \frac{5}{3}x^3 \right)
\\
&= 2^{-\frac{2}{3}} - 2^{-\frac{2}{3}}x + \frac{5 \cdot 2^{-\frac{2}{3}}}{4}x^2 - \frac{5 \cdot 2^{-\frac{2}{3}}}{3}x^3.
\end{align*}
\]
</li>
</ol>]]></text>
    </questiontext>
    <generalfeedback format="moodle_auto_format">
      <text/>
    </generalfeedback>
    <defaultgrade>0</defaultgrade>
    <penalty>0</penalty>
    <hidden>0</hidden>
    <idnumber/>
  </question>
</quiz>
