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

Dec 8, 2016

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

Explanation:

The binomial series is

${\left(a + b\right)}^{n} = {a}^{n} + n {a}^{n - 1} b + \frac{n \left(n - 1\right)}{1 \cdot 2} {a}^{n - 2} {b}^{2} + \frac{n \left(n - 1\right) \left(n - 2\right)}{1 \cdot 2 \cdot 3} {a}^{n - 3} {b}^{3} + \frac{n \left(n - 1\right) \left(n - 2\right) \left(n - 3\right)}{1 \cdot 2 \cdot 3 \cdot 4} {a}^{n - 4} {b}^{4} + \ldots .$

In our case, we have ${\left(1 - 2 x\right)}^{\frac{1}{3}}$

$a = 1$

$b = - 2 x$

$n = \frac{1}{3}$

So,

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

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