# How do you use the binomial series to expand (1+9x)^(1/3)?

Jan 31, 2017

${\left(1 + 9 x\right)}^{\frac{1}{3}} = 1 + 3 x - 9 {x}^{2} + 45 {x}^{3} - 270 {x}^{4} + \ldots \ldots .$

#### Explanation:

According to binomial series expansion
(1+a)^n=1+na+(n(n-1))/(2!)a^2+(n(n-1)(n-2))/(3!)a^3+(n(n-1)(n-2)(n-3))/(4!)a^4+.......

Hence (1+9x)^(1/3)=1+(1/3)(9x)+((1/3)(1/3-1))/(2!)(9x)^2+((1/3)(1/3-1)(1/3-2))/(3!)(9x)^3+((1/3)(1/3-1)(1/3-2)(1/3-3))/(4!)(9x)^4+.......

or ${\left(1 + 9 x\right)}^{\frac{1}{3}} = 1 + 3 x + \frac{\left(\frac{1}{3}\right) \left(- \frac{2}{3}\right)}{2} 81 {x}^{2} + \frac{\left(\frac{1}{3}\right) \left(- \frac{2}{3}\right) \left(- \frac{5}{3}\right)}{6} 729 {x}^{3} + \frac{\left(\frac{1}{3}\right) \left(- \frac{2}{3}\right) \left(- \frac{5}{3}\right) \left(- \frac{8}{3}\right)}{24} 6561 {x}^{3} + \ldots \ldots .$

or ${\left(1 + 9 x\right)}^{\frac{1}{3}} = 1 + 3 x - 9 {x}^{2} + 45 {x}^{3} - 270 {x}^{4} + \ldots \ldots .$