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

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.
