Alien's fingers

Let $k=0.1212\ldots$ .

$r^2k=12.1212\ldots$

$(r^2-1)k=12=r+2$

@Shaun Leong is right. Let me explain the same thing in my own way for you to have more than one possible explanation of it. Since $12_r=r+2,$ then $(0.121212...)_r= 12_r (\frac{1}{r^2}+\frac{1}{r^4}+...)=(r+2)\frac{1}{r^2}(1+\frac{1}{r^2}+\frac{1}{r^4}+...)=(r+2)\frac{1}{r^2}(\frac{1}{1-\frac{1}{r^2}})=\frac{r+2}{r^2-1}.$

OK, a number between 0 and 1 in base 10 can be expressed in the form $0.a_1 a_2 a_3\dots a_n \dots$ in base 10. For example, $34/99=0.343434\dots.$ But what is the meaning of this notation? By definition $0.343434...=\frac{3}{10}+\frac{4}{10^2}+\frac{3}{10^3}+\frac{4}{10^4}+\dots,$ when the base is 10. By combining the first term with the second, the third with the four and so for, we obtain that the previous number can also be written as $\frac{34}{100}+\frac{34}{100^2}+\dots$ When you use a base $r$ that can be any positive integer (not necessarily 10), then , for example, $(0.121212\dots)_r=\frac{1}{r}+\frac{2}{r^2}+\frac{1}{r^3}+\frac{2}{r^4}+\dots$ Combining the first with the second term, the third with the fourth, etc, we obtain that $(0.121212\dots)_r= (\frac{1}{r}+\frac{2}{r^2})+(\frac{1}{r^3}+\frac{2}{r^4})+\dots=\frac{r+2}{r^2}+\frac{r+2}{r^4}+\dots= \frac{12_r}{r^2}+\frac{12_r}{r^4}+\dots$