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Expert-verified Found in: Page 680 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # In exercises 59-62 concern the binomial series to find the maclaurin series for the given function .$\left(1+x{\right)}^{2/3}$

The maclaurin series for the given function is

$\left(1+x{\right)}^{\frac{2}{3}}=1+\frac{2}{3}x-\frac{1}{9}{x}^{2}+\frac{4}{81}{x}^{3}+\cdots$

See the step by step solution

## Step 1. Given information

We have been given

$\left(1+x{\right)}^{2/3}$

to find the maclaurin series by using binomial series

## Step 2.Defining the series

For any non- zero constant p, the Maclaurin series for the function $g\left(x\right)=\left(1+x{\right)}^{p}$

is called the binomial series which is given by $\sum _{k=0}^{\mathrm{\infty }} \left(\begin{array}{l}p\\ k\end{array}\right){x}^{k}$

where the binomial coefficient is

$\left(\begin{array}{l}p\\ k\end{array}\right)=\left\{\begin{array}{r}\frac{p\left(p-1\right)\left(p-2\right)\cdots \left(p-k+1\right)}{k!}, \text{if}k>0\\ 1, \text{if}k=0\end{array}\right\$

## Step 3. Binomial series for the given function is

So for the given function $f\left(x\right)=\left(1+x{\right)}^{\frac{2}{3}}$ binomial series is ,

$\left(1+x{\right)}^{\frac{2}{3}}=\sum _{k=0}^{\mathrm{\infty }} \left(\begin{array}{l}\frac{2}{3}\\ k\end{array}\right){x}^{k}$

implies that ,

$\begin{array}{r}\left(1+x{\right)}^{\frac{2}{3}}=\left(\begin{array}{c}\frac{2}{3}\\ 0\end{array}\right){x}^{0}+\left(\begin{array}{c}\frac{2}{3}\\ 1\end{array}\right){x}^{1}+\left(\begin{array}{c}\frac{2}{3}\\ 2\end{array}\right){x}^{2}+\left(\begin{array}{c}\frac{2}{3}\\ 3\end{array}\right){x}^{3}+\cdots \\ =1+\frac{2}{3}x+\frac{\frac{2}{3}\left(-\frac{1}{3}\right)}{2!}{x}^{2}+\frac{\frac{2}{3}\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{3!}{x}^{3}+\cdots \\ =1+\frac{2}{3}x-\frac{1}{9}{x}^{2}+\frac{4}{81}{x}^{3}+\cdots \end{array}$

## Step 4. The maclaurin series for given function is

The maclaurin series for the given function is $\left(1+x{\right)}^{\frac{2}{3}}=1+\frac{2}{3}x-\frac{1}{9}{x}^{2}+\frac{4}{81}{x}^{3}+\cdots$ ### Want to see more solutions like these? 