Does this pattern continue $\lfloor\sqrt{44}\rfloor=6, \lfloor\sqrt{4444}\rfloor=66,\dots$?

Solution 1:

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.

Solution 2:

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: $66...66 = \frac{2}{3} 10^n - \frac{2}{3} < \frac{2}{3} \sqrt{10^{2n} - 1} < \frac{2}{3} 10^n = 66...66.67$.