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

Nov 19, 2016

(1+x^2)^(2/3)=sum_(k=0)^oo(Pi_(j = 1)^k(2/3 - j + 1))/(k!)x^(2k) convergent for $\left\mid x \right\mid < 1$
(1+a)^n=1+na+(n(n-1))/(2!)a^2+cdots+(Pi_(j=k)^n (n-j+1))/(k!)a^k
(1+x^2)^(2/3)=sum_(k=0)^oo(Pi_(j = 1)^k(2/3 - j + 1))/(k!)x^(2k) convergent for $\left\mid x \right\mid < 1$