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

Nov 16, 2016

The series is $= - 1 + \frac{2 x}{3} + \frac{4 {x}^{2}}{9} + \frac{40 {x}^{2}}{81} + \ldots$

#### Explanation:

${\left(a + b\right)}^{n} = {a}^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {a}^{n - 1} b + \left(\begin{matrix}n \\ 2\end{matrix}\right) {a}^{n - 2} {b}^{2} + \left(\begin{matrix}n \\ 3\end{matrix}\right) {a}^{n - 2} {b}^{3} + . .$

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

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

We rewrite the expression as ${\left(2 x - 1\right)}^{\frac{1}{3}} = {\left(- 1 + 2 x\right)}^{\frac{1}{3}}$

$= {\left(- 1\right)}^{\frac{1}{3}}$ + $\frac{1}{3} \cdot \left({\left(- 1\right)}^{- \frac{2}{3}}\right) \cdot \left(2 x\right)$ +  (1/3) (-2/3)) * (1/2)* (-1)^(-5/6) ( 2 x)^2  + $\left(\frac{1}{3}\right) \left(- \frac{2}{3}\right) \left(- \frac{5}{3}\right) \cdot \left(\frac{1}{6}\right) {\left(2 x\right)}^{3}$

$= - 1 + \frac{2 x}{3} + \frac{4 {x}^{2}}{9} + \frac{40 {x}^{2}}{81} + \ldots$