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In exercises 59-62 concern the binomial series to find the maclaurin series for the given function .

(1+x)2/3

Short Answer

Expert verified

The maclaurin series for the given function is

(1+x)23=1+23x−19x2+481x3+⋯

Step by step solution

01

Step 1. Given information 

We have been given

(1+x)2/3

to find the maclaurin series by using binomial series

02

Step 2.Defining the series 

For any non- zero constant p, the Maclaurin series for the function g(x)=(1+x)p

is called the binomial series which is given by ∑k=0∞ pkxk

where the binomial coefficient is

pk=p(p−1)(p−2)⋯(p−k+1)k!, â¶Ä…â¶Ä…â¶Ä…ifk>01, â¶Ä…â¶Ä…â¶Ä…ifk=0

03

Step 3. Binomial series for the given function is 

So for the given functionf(x)=(1+x)23binomial series is ,

(1+x)23=∑k=0∞ 23kxk

implies that ,

(1+x)23=230x0+231x1+232x2+233x3+⋯=1+23x+23−132!x2+23−13−433!x3+⋯=1+23x−19x2+481x3+⋯

04

Step 4. The maclaurin series for given function is 

The maclaurin series for the given function is(1+x)23=1+23x−19x2+481x3+⋯

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