Question

find the first four non-zero terms in the binomial expansion of
$(1 + \frac{x}{125} )^{ \frac{1}{3} }$
in ascending powers of
$x$

Answer 1

$\displaystyle \left(1+\frac{x}{125}\right)^{\frac{1}{3}}=\sum\limits_{n=0}^{\infty}\binom{\frac{1}{3}}{n}\left(\frac{x}{125}\right)^{n}\\\\ \approx 1+\frac{\left(\frac{1}{3}\right)}{1!}\left(\frac{x}{125}\right)+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}\left(\frac{x}{125}\right)^{2}+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}\left(\frac{x}{125}\right)^{3}\\\\ \approx 1+\frac{1}{375}x-\frac{1}{140625}x^{2}+\frac{1}{31640625}x^{3}$