Expanding $\frac{1}{\sqrt{1-x^2}}$ through the expansion of $\frac{1}{\sqrt{1-x}}$ by binomial series

Question

I heard that to expand $\frac{1}{\sqrt{1-x^2}}$, I have to expand $\frac{1}{\sqrt{1-x}}$ by binomial series and then just replace $x$ to $x^2$. Using binomial series, I found that $\frac{1}{\sqrt{1-x}}$ becomes $1+\sum_{n=1}^{\infty}\frac{1*3*5*....*(2n-1)}{2^nn!}(1-x)^{-\frac{2n+1}{2}}$. So is $ 1+\sum_{n=1}^{\infty}\frac{1*3*5*....*(2n-1)}{2^nn!}(1-x^2)^{-\frac{2n+1}{2}}$ would be an right binomial expansion of $\frac{1}{\sqrt{1-x^2}}$?

Answer 1

rewrite the function as $(1-x^2)^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k}(-x)^{2k}$. Now expand $\binom{-\frac{1}{2}}{k}$, you'll get something qutie similar to you expression in the second line. This will be your solution.