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

1 Answer
Sep 14, 2016

Answer:

#(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+....................#