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      <text>16.5.3.2 Differentiate a Maclaurin series-a</text>
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      <text><![CDATA[<p class="HELM_exercise">Exercise</p>
<p>Using differentiation, find the Maclaurin expansion of <span class="math inline">\((1-x)^{-3}\)</span>.</p><p>(<em>Hint&nbsp;</em>: \(\dfrac{1}{1-x}=1+x+x^2+\cdots\qquad |x|&lt;1\))</p>]]></text>
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      <text><![CDATA[<p dir="ltr" style="text-align: left;"><br></p>
<p>Now, by differentiation, obtain the expansion of <span class="math inline">\(\dfrac{1}{(1-x)^2}\)</span>: <span class="math display">\[\begin{aligned} \dfrac{1}{(1-x)^2}=\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{1}{1-x}\right) &amp;= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(1+x+x^2+\cdots\right), \\ &amp;= 1+2x+3x^2+4x^3 \end{aligned}\]</span> Differentiate again to obtain the expansion of <span class="math inline">\((1-x)^{-3}\)</span>: <span class="math display">\[\begin{aligned} \dfrac{1}{(1-x)^3}=\dfrac{1}{2}\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{1}{(1-x)^2}\right) &amp;= \dfrac{1}{2}\left[2+6x+12x^2+20x^3+\cdots\right], \\ &amp;= 1+3x+6x^2+10x^3+\cdots \end{aligned}\]</span></p>]]></text>
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    <answer fraction="100" format="html">
      <text><![CDATA[<p dir="ltr" style="text-align: left;">\(1+3x+6x^2+10x^3+\cdots\)<br></p>]]></text>
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    <answer fraction="0" format="html">
      <text><![CDATA[<p dir="ltr" style="text-align: left;">\(2+6x+12x^2+20x^3+\cdots\)<br></p>]]></text>
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    <answer fraction="0" format="html">
      <text><![CDATA[<p dir="ltr" style="text-align: left;">\(1+2x+3x^2+4x^3+\cdots\)<br></p>]]></text>
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