how can I prove that $\frac{1}{\sqrt{1-x^2}}=\sum _{n=0}^{\infty }\frac{(2n)!}{4^n(n!)^2}x^{2n}$?
I tried using Newton's generalized binomial theorem to get:$$\frac{1}{\sqrt{1-x^2}}=(1-x^2)^{-1/2}=\sum_{k=0}^{\infty}{\begin{pmatrix} -1/2\\ k \end{pmatrix}}(-x^2)^k=\sum_{k=0}^{\infty}{(-1)^k\begin{pmatrix} -1/2\\ k \end{pmatrix}}x^{2k}$$ How is that equal to to $\sum _{n=0}^{\infty }\frac{(2n)!}{4^n(n!)^2}x^{2n}$?
Well, you just need to go one step further. Detailed steps are
$$ (-1)^k\binom{-1/2}{k} x^{2k} = (-1)^k\frac{\prod_{i=1}^k(-1/2-i+1)}{\prod_{i=1}^k i}x^{2k} $$ $$ = \frac{\prod_{i=1}^k(1/2+i-1)}{\prod_{i=1}^k i }x^{2k} = \frac{\prod_{i=1}^k(2i-1)}{2^k\prod_{i=1}^k i }x^{2k} $$ $$ = \frac{\prod_{i=1}^k(2i-1)\prod_{j=1}^k j}{2^k\prod_{i=1}^k i \prod_{j=1}^k j }x^{2k} = \frac{\prod_{i=1}^k(2i-1)\prod_{j=1}^k (2j)}{4^k{\left(\prod_{i=1}^k i\right)}^2 }x^{2k} $$ $$ = \frac{\prod_{i=1}^{2k}i}{4^k{\left(\prod_{i=1}^k i\right)}^2 }x^{2k} = \frac{(2k)!}{4^k{\left(k!\right)}^2 }x^{2k} $$