Answer:
First term=1
Second term=-x
Third term=[tex]2x^2[/tex]
Fourth term =[tex]-\frac{28}{3!}x^3[/tex]
Step-by-step explanation:
We are given that function
[tex]f(x)=(1+3x)^{-1/3}[/tex]
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
[tex]f(x)=f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+....[/tex]
[tex]f(0)=(1+3x)^{\frac{-1}{3}}=1[/tex]
[tex]f'(x)=-\frac{1}{3}(1+3x)^{-\frac{4}{3}}(3)=-(1+3x)^{-\frac{4}{3}}[/tex]
[tex]f'(0)=-1[/tex]
[tex]f''(x)=\frac{4}{3}\times 3 (1+3x)^{-\frac{7}{3}}[/tex]
[tex]f''(0)=4[/tex]
[tex]f'''(x)=-4\times \frac{7}{3}\times 3(1+3x)^{-\frac{10}{3}}[/tex]
[tex]f'''(0)=-28[/tex]
Substitute the values we get
[tex](1+3x)^{-\frac{1}{3}}=1-x+\frac{4}{2!}x^2+\frac{-28}{3!}x^3+...[/tex]
[tex](1+3x)^{-\frac{1}{3}}=1-x+2x^2+\frac{-28}{3!}x^3+...[/tex]
First term=1
Second term=-x
Third term=[tex]2x^2[/tex]
Fourth term =[tex]-\frac{28}{3!}x^3[/tex]
