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

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Answer 1 · 2016-11-16T12:53:30

The series is #=-1+(2x)/3+(4x^2)/9+(40x^2)/81+...#

#(a+b)^n=a^n+((n),(1))a^(n-1)b+((n),(2))a^(n-2)b^2+((n),(3))a^(n-2)b^3+..#

#((n),(1))=(n!)/((n-1)!1!)=n#

#((n),(2))=(n!)/((n-2)!(2!))=(n(n-1))/(1*2)#

We rewrite the expression as #(2x-1)^(1/3)=(-1+2x)^(1/3)#

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

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