use Binomial THM to show that:
$\frac{1}{\sqrt{1-4x}}$=$\sum\limits_{m=0}^\infty {2m \choose m} x^m$
Also, what is the interval of convergence of this power series?
What I tried:
Since the bino. tells us that: $(1+x)^n$=$\sum\limits_{k=0}^n {n \choose k} x^k$
then I tried to start writing out the series, by replacing $n$ with $2m$ and $k$ with $m$
${2m \choose 0} x^0$+${2m \choose 1} x^1$+${2m \choose 2} x^2$+...+${2m \choose 2m-1} x^{2m-1}$+${2m \choose 2m} x^{2m}$
this should give me:
${0 \choose 0} x^0$+${2 \choose 1} x^1$+${4 \choose 2} x^2$+...+${2m \choose 2m-1} x^{2m-1}$+${2m \choose 2m} x^{2m}$
simplify to:
1+$2x$+$6x^2$+$20x^3$+....+ $x^{2m}$
but now I get stuck :( Please help! Thank you!
I think that you have to use the more general form of the binomial theorem: for any $\alpha, y\in\mathbb{R}$ such that $\vert y\vert<1$, we have $$ (1+y)^\alpha=\sum_{r=0}^\infty {\alpha \choose r}y^r $$ Now, set $y=-4x$ and $\alpha=-1/2$ to get $$ \frac{1}{\sqrt{1-4x}}=\sum_{r=0}^\infty {-\frac{1}{2}\choose r}(-4x)^r $$ So we only have to see that $ {-\frac{1}{2}\choose r}(-4x)^r= {2r\choose r}x^r$.
By definition, $$ {-\frac{1}{2}\choose r}(-4x)^r= \frac{(-1/2)\cdots (-1/2-(r-1))}{r!}(-4)^rx^r=\frac{2^r(2r-1)(2r-3)\cdots 1}{r!}x^r $$ Note that $$ 2^r\frac{(2r-1)(2r-3)\cdots 1}{r!}\frac{r!}{r!}= \frac{(2r-1)\cdots 1}{r!}\frac{2r(2r-2)\cdots 2}{r!}=\frac{(2r)!}{r!r!}={2r \choose r} $$ Therefore, ${-\frac{1}{2}\choose r}(-4x)^r={2r \choose r}x^r$ and the result follows.