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a_sh-v [17]
2 years ago
7

Use the appropriate substitutions to write down the first four nonzero terms of the Maclaurin series for the binomial (1+3x)^(-1

/3)
Mathematics
1 answer:
PolarNik [594]2 years ago
5 0

Answer:

First term=1

Second term=-x

Third term=2x^2

Fourth term =-\frac{28}{3!}x^3

Step-by-step explanation:

We are given that function

f(x)=(1+3x)^{-1/3}

We have to find the  first four non zero terms of the Maclaurin series for the binomial.

Maclaurin series of function f(x) is given by

f(x)=f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+....

f(0)=(1+3x)^{\frac{-1}{3}}=1

f'(x)=-\frac{1}{3}(1+3x)^{-\frac{4}{3}}(3)=-(1+3x)^{-\frac{4}{3}}

f'(0)=-1

f''(x)=\frac{4}{3}\times 3 (1+3x)^{-\frac{7}{3}}

f''(0)=4

f'''(x)=-4\times \frac{7}{3}\times 3(1+3x)^{-\frac{10}{3}}

f'''(0)=-28

Substitute the values we get

(1+3x)^{-\frac{1}{3}}=1-x+\frac{4}{2!}x^2+\frac{-28}{3!}x^3+...

(1+3x)^{-\frac{1}{3}}=1-x+2x^2+\frac{-28}{3!}x^3+...

First term=1

Second term=-x

Third term=2x^2

Fourth term =-\frac{28}{3!}x^3

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Answer:

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Step-by-step explanation:

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alex41 [277]

Answer:

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Step-by-step explanation:

Hope that this helps you out! :)  

If you have any questions please put them in the comment section below this answer.  

Have a great rest of your day/night!  

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2 years ago
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