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Section 10.18 : Binomial Series
4. Write down the first four terms in the binomial series for \(\sqrt[3]{{8 - 2x}}\).
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Start SolutionFirst, we need to make sure it is in the proper form to use the Binomial Series. Here is the proper form for this function,
\[\sqrt[3]{{8 - 2x}} = {\left( {8\left( {1 - \frac{1}{4}x} \right)} \right)^{\frac{1}{3}}} = {\left( 8 \right)^{\frac{1}{3}}}{\left( {1 - \frac{1}{4}x} \right)^{\frac{1}{3}}} = 2{\left( {1 + \left( { - \frac{1}{4}x} \right)} \right)^{\frac{1}{3}}}\]Recall that for proper from we need it to be in the form “1+” and so we needed to factor the 8 out of the root and “move” the minus sign into the second term. Also, as we can see we will have \(k = \frac{1}{3}\)
Show Step 2Now all we need to do is plug into the formula from the notes and write down the first four terms.
\[\begin{align*}\sqrt[3]{{8 - 2x}} & = 2{\left( {1 + \left( { - \frac{1}{4}x} \right)} \right)^{\frac{1}{3}}}\\ & = 2\sum\limits_{i = 0}^\infty { {\frac{1}{3} \choose i} {{\left( { - \frac{1}{4}x} \right)}^i}} \\ & = 2\left[ {1 + \left( {\frac{1}{3}} \right){{\left( { - \frac{1}{4}x} \right)}^1} + \frac{{\left( {\frac{1}{3}} \right)\left( { - \frac{2}{3}} \right)}}{{2!}}{{\left( { - \frac{1}{4}x} \right)}^2} + \frac{{\left( {\frac{1}{3}} \right)\left( { - \frac{2}{3}} \right)\left( { - \frac{5}{3}} \right)}}{{3!}}{{\left( { - \frac{1}{4}x} \right)}^3} + \cdots } \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{2 - \frac{1}{6}x - \frac{1}{{72}}{x^2} - \frac{5}{{2592}}{x^3} + \cdots }}\end{align*}\]