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

Question

11379 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-14T13:16:14

$(1+x^2)^(-1/3)=1-1/3x^2+2/9x^4-14/81x^6+35/243x^8+.......$

According to binomial series $(1+a)^n=1+na+(n(n-1))/(2!)a^2+(n(n-1)(n-2))/(3!)a^3+(n(n-1)(n-2)(n-3))/(4!)a^4+.......$

Hence $(1+x^2)^(-1/3)=1+(-1/3)x^2+((-1/3)(-1/3-1))/(2!)x^4+((-1/3)(-1/3-1)(-1/3-2))/(3!)x^6+((-1/3)(-1/3-1)(-1/3-2)(-1/3-3))/(4!)x^8+.......$

or $(1+x^2)^(-1/3)=1-1/3x^2+((-1/3)(-4/3))/(2!)x^4+((-1/3)(-4/3)(-7/3))/(3!)x^6+((-1/3)(-4/3)(-7/3)(-10/3))/(4!)x^8+.......$

or $(1+x^2)^(-1/3)=1-1/3x^2+4/18x^4-28/162x^6+280/1944x^8+.......$

or $(1+x^2)^(-1/3)=1-1/3x^2+2/9x^4-14/81x^6+35/243x^8+.......$

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