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

Feb 20, 2017

The answer is $= 1 + \frac{1}{3} {x}^{2} - \frac{1}{9} {x}^{4} + \frac{5}{162} {x}^{6} + \ldots \ldots$

#### Explanation:

The binomial series is

${\left(1 + b\right)}^{\alpha} = {\sum}_{k = 0}^{\infty} \left(\begin{matrix}\alpha \\ k\end{matrix}\right) {b}^{k}$

=1+alphab+(alpha(alpha-1))/(2!)b^2+(alpha(alpha-1)(alpha-2))/(3!)b^3+.......

Here,

we have

$b = {x}^{2}$

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

${\left(1 + {x}^{2}\right)}^{\frac{1}{3}} = 1 + \frac{1}{3} {x}^{2} + \frac{\frac{1}{3} \cdot - \frac{2}{3}}{1 \cdot 2} {x}^{4} + \frac{\frac{1}{3} \cdot - \frac{2}{3} \cdot - \frac{5}{6}}{1 \cdot 2 \cdot 3} {x}^{6} + . .$

$= 1 + \frac{1}{3} {x}^{2} - \frac{1}{9} {x}^{4} + \frac{5}{162} {x}^{6} + \ldots \ldots$