How do you use the binomial series to expand #(1-2x)^(1/3)#?

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Answer 1 · 2016-12-08T15:58:21

The answer is #=1-2/3x-4/9x^2-40/81x^3-....#

The binomial series is

#(a+b)^n=a^n+na^(n-1)b+(n(n-1))/(1*2)a^(n-2)b^2+(n(n-1)(n-2))/(1*2*3)a^(n-3)b^3+(n(n-1)(n-2)(n-3))/(1*2*3*4)a^(n-4)b^4+....#

In our case, we have #(1-2x)^(1/3)#

#a=1#

#b=-2x#

#n=1/3#

So,

#(1-2x)^(1/3)=1-1/3*2x-1/3*2/3*1/2(-2x)^2-1/3*2/3*5/3*1/6(-2x)^3-...#

#=1-2/3x-4/9x^2-40/81x^3-....#