Question

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

Answer 1

`(1+x^2)^(2/3)=sum_(k=0)^oo(Pi_(j = 1)^k(2/3 - j + 1))/(k!)x^(2k)` convergent for `absx lt 1`

Explanation:

`(1+a)^n=1+na+(n(n-1))/(2!)a^2+cdots+(Pi_(j=k)^n (n-j+1))/(k!)a^k`

so

`(1+x^2)^(2/3)=sum_(k=0)^oo(Pi_(j = 1)^k(2/3 - j + 1))/(k!)x^(2k)` convergent for `absx lt 1`