By observing the following I have a feeling that the pattern continues.
$$\lfloor \sqrt{44} \rfloor=6$$ $$\lfloor \sqrt{4444} \rfloor=66$$ $$\lfloor \sqrt{444444} \rfloor=666$$ $$\lfloor \sqrt{44444444} \rfloor=6666$$
But I'm unable to prove it. Your help will be appreciated.
On
As a somewhat inferior alternative to Peter's algebraic approach, you could work out a square root in long-division style as shown here:
\begin{array}{r} 6 \; \phantom{0}6 \;\phantom{0}6 \;\phantom{0}6 \\[-3pt] \sqrt{44 \; 44 \; 44 \; 44} \\[-3pt] \underline{36}\; \phantom{66 \; 00 \; 00} \\[-3pt] 12\underline6|\phantom{0}8 \; 44 \phantom{\; 00 \; 00}\\[-3pt] \underline{7 \; 56} \phantom{\; 00 \; 00}\\[-3pt] 132\underline{6} |\phantom{0} \; 88 \; 44 \phantom{\; 00}\\[-3pt] \underline{79 \; 56} \phantom{\; 00}\\[-3pt] 1332\underline{6} | \phantom{0}8 \; 88 \;44\\[-3pt] \underline{7 \; 99 \; 56} \end{array}
From here you may be able to extrapolate some provable patterns that occur as you append more pairs of $4$s to the digits under the square root sign.
On
This can be proved a bit more simple.
The general term can be written as $4 \frac{10^{2n} - 1}{9}$. Taking square root will give you $\frac{2}{3} \sqrt{10^{2n} - 1}$. As $\frac{2}{3} \approx 0.66666666666$ and $\sqrt{10^{2n} - 1} \approx \sqrt{10^{2n}} = 10^n$, this explains why the result is $6;66;666;...$ etc. To prove rigorously, observe that: $66...66 = \frac{2}{3} 10^n - \frac{2}{3} < \frac{2}{3} \sqrt{10^{2n} - 1} < \frac{2}{3} 10^n = 66...66.67$.
Hint :
We have $$\left(\frac{6\cdot (10^n-1)}{9}\right)^2=\frac{4\cdot (10^{2n}-1)}{9}-\frac{8\cdot (10^n-1)}{9}$$
Try to find out why this proves that the pattern continues forever.