Question

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

Answer 1

`(1+9x)^(1/3)=1+3x-9x^2+45x^3-270x^4+.......`

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 `(1+9x)^(1/3)=1+3x+((1/3)(-2/3))/(2)81x^2+((1/3)(-2/3)(-5/3))/(6)729x^3+((1/3)(-2/3)(-5/3)(-8/3))/(24)6561x^3+.......`

or `(1+9x)^(1/3)=1+3x-9x^2+45x^3-270x^4+.......`