Question

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

Answer 1

`(1-x)^(1/3)=1-1/3x-1/9x^2-5/81x^3-10/243x^12+....................`

Explanation:

Binomial theorem gives the expansion of `(1+x)^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 `(1-x)^(1/3)`

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

= `1-1/3x+((1/3)(-2/3))/(2!)x^2-((1/3)(-2/3)(-5/3))/(3!)x^3+((1/3)(-2/3)(-5/3)(-8/3))/(4!)x^4+....................`

= `1-1/3x-1/9x^2-5/81x^3-10/243x^12+....................`