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

Sep 13, 2016

$\frac{1}{1 + {x}^{3}} ^ \left(\frac{1}{2}\right) = 1 - \frac{1}{2} {x}^{3} + \frac{3}{8} {x}^{6} - \frac{15}{48} {x}^{9} + \frac{105}{384} {x}^{12} + \ldots \ldots \ldots \ldots$

#### Explanation:

Binomial theorem gives the expansion of ${\left(1 + x\right)}^{n}$ as

(1+x)^n=1+nx+(n(n-1))/(2!)x^2+(n(n-1(n-2)))/(3!)x^3+(n(n-1)(n-2)(n-3))/(4!)x^4+....................

Hence $\frac{1}{1 + {x}^{3}} ^ \left(\frac{1}{2}\right) = {\left(1 + {x}^{3}\right)}^{- \frac{1}{2}}$

= 1+(-1/2)x^3+((-1/2)(-1/2-1))/(2!)x^6+((-1/2)(-1/2-1)(-1/2-2))/(3!)x^9+((-1/2)(-1/2-1)(-1/2-2)(-1/2-3))/(4!)x^12+....................

= 1-1/2x^3+((-1/2)(-3/2))/(2!)x^6+((-1/2)(-3/2)(-5/2))/(3!)x^9+((-1/2)(-3/2)(-5/2)(-7/2))/(4!)x^12+....................

= $1 - \frac{1}{2} {x}^{3} + \frac{3}{8} {x}^{6} - \frac{15}{48} {x}^{9} + \frac{105}{384} {x}^{12} + \ldots \ldots \ldots \ldots \ldots \ldots . .$