value of $k$ in binomial sum 
If $\displaystyle (1-x)^{\frac{1}{2}}=a_{0}+a_{1}x+a_{2}x^2+\cdots \cdots +\infty.$ and $a_{0}+a_{1}+a_{2}+\cdots+a_{10}=\frac{\binom{20}{10}}{k^{10}}.$ Then $k$ is 

what i try
$(1-x)^{\frac{1}{2}}=1-\frac{1}{2}x+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\frac{x^2}{2!}+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\bigg(\frac{1}{2}-2\bigg)\frac{x^3}{3!}+\cdots\cdots$
$\displaystyle a_{0}=1,a_{1}=-\frac{1}{2},a_{2}=-\frac{1}{2^2}\frac{1}{2!},a_{3}=\frac{1\cdot 3}{2^3}\frac{1}{3!}+\cdots$
How do i solve it Help me pleses
 A: We have
$$\sum_{n=0}^{1}(-1)^n\binom{\frac 12}{n}=\frac 12=\frac 24=\frac{\binom{2}{1}}{4^1},\quad \sum_{n=0}^{2}(-1)^n\binom{\frac 12}{n}=\frac 38=\frac{6}{16}=\frac{\binom{4}{2}}{4^2}$$
$$\sum_{n=0}^{3}(-1)^n\binom{\frac 12}{n}=\frac{5}{16}=\frac{20}{64}=\frac{\binom{6}{3}}{4^3},\quad \sum_{n=0}^{4}(-1)^n\binom{\frac 12}{n}=\frac{35}{128}=\frac{70}{256}=\frac{\binom{8}{4}}{4^4}$$
So, it seems that the following holds :
$$\sum_{n=0}^{r}(-1)^n\binom{\frac 12}{n}=\frac{\binom{2r}{r}}{4^r}\tag1$$
One can prove $(1)$ by induction.
Proof : 
$(1)$ holds for $r=0$.
Supposing that $(1)$ holds for $r$ gives
$$\begin{align}\sum_{n=0}^{r+1}(-1)^n\binom{\frac 12}{n}&=\sum_{n=0}^{r}(-1)^n\binom{\frac 12}{n}+(-1)^{r+1}\binom{\frac 12}{r+1}
\\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{\frac 12(\frac 12-1)\cdots (\frac 12-r)}{(r+1)!}
\\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{(2r-1)(2r-3)\cdots (2-1)(-1)}{(r+1)!(-2)^{r+1}}
\\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{-(2r-1)(2r-3)\cdots (2-1)\cdot (2r)(2r-2)\cdots 2}{(r+1)!(-2)^{r+1}\cdot (2r)(2r-2)\cdots 2}
\\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{-(2r)!}{(r+1)!(-2)^{r+1}\cdot 2^rr!}
\\\\&=\frac{\binom{2r}{r}}{4^r}+\frac{-1}{(r+1)\cdot 2^{2r+1}}\cdot\frac{(2r)!}{r!r!}
\\\\&=\frac{\binom{2r}{r}}{4^r}\bigg(1-\frac{1}{2(r+1)}\bigg)
\\\\&=\frac{1}{4^r}\cdot\frac{(2r)!}{r!r!}\cdot\frac{2r+1}{2r+2}\cdot\frac{2r+2}{2r+2}
\\\\&=\frac{\binom{2r+2}{r+1}}{4^{r+1}}\qquad\square\end{align}$$
It follows from $(1)$ that
$$a_0+a_1+\cdots +a_{10}=\sum_{n=0}^{10}(-1)^n\binom{\frac 12}{n}=\frac{\binom{20}{10}}{\color{red}4^{10}}$$
A: The binomial expansion is
$$\sqrt{1-x}=\sum_{n=0}^\infty (-1)^n\binom{\frac{1}{2}}{n}x^n$$ and so you need to solve for $k$
$$\sum_{n=0}^{10} (-1)^n\binom{\frac{1}{2}}{n}=\frac{\binom{20}{10}}{k^{10}}$$ The result should be a rather small integer. Just use Excel and inspection.
