At least one even number among $\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$ 
For any positive integer $n$, prove that the set
$$\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$$
contains at least one even number.

I tried to prove this question by contradiction, assuming that each element is odd. There exist the positive integers $k_1, k_2, ..., k_{n+1}$ such that
$$2k_1-1<2^n\sqrt{2}<2k_1$$
$$2k_2-1<2^{n+1}\sqrt{2}<2k_2$$
$$...$$
$$2k_{n+1}-1<2^{2n}\sqrt{2}<2k_{n+1}$$
But i can't find a contradiction among these inequalities.

Let me continue TonyK's comment. If $\sqrt{2}=1.b_1b_2\ldots b_{n-1}11\ldots11b_{2n+1}\ldots$, then 
$$
\sqrt{2}=\frac{k}{2^{n-1}}+\left(\frac{1}{2^n}+\ldots+\frac{1}{2^{2n}}\right)+r,
$$
where $k=(\overline{1b_1b_2\ldots b_{n-1}})_2$ is a positive integer and $r\in(0,\frac{1}{2^{2n}})$ (since $\sqrt{2}$ is not rational number of the form $p/2^{q}$). Hence,
$$
2^{n+1}(k+1)-1<2^{2n}\sqrt{2}<2^{n+1}(k+1).
$$
Denote $m=2^{n+1}(k+1)$, then the last inequality can be rewritten as follows
$$
(m-1)^2<2^{4n+1}<m^2.
$$
However, $m^2$ and $2^{4n+1}$ are divisible by $2^{2n+2}$, so 
$$
(m-1)^2\leq 2^{4n+1}\leq m^2-2^{2n+2}.
$$
Thus, $2m-1\geq 2^{2n+2}$, so $m>2^{2n+1}$. Recall that $m=2^{n+1}(k+1)$, so the last inequality means that $k\geq 2^{n}$. But it's impossible because $k=(\overline{1b_1b_2\ldots b_{n-1}})_2<2^n$.
