There are two such possible strings, 0 and 1, so [tex]S_1=2[/tex].
1 can be appended to either string, while 0 can only be appended to one, so [tex]S_2=3[/tex].
For an [tex](n+1)[/tex]-string to end with a 0, the bit in the [tex]n[/tex]th position needs to be a 1 - by definition, there are [tex]S_n[/tex] such possible strings.
If an [tex]n[/tex]-string already ends in a 0, we can drop this last bit to get an [tex](n-1)[/tex]-string that ends in a 1 (we know this occurs because 00 is not allowed) - by definition there are [tex]S_{n-1}[/tex] such strings - and append 10 to it instead, giving an [tex](n+1)[/tex]-string that ends in a 0.
So, the number of [tex]n[/tex] bit strings not containing consecutive 0s is given recursively by
[tex]\begin{cases}S_1=2\\S_2=3\\S_{n+1}=S_n+S_{n-1}&\text{for }n\ge2\end{cases}[/tex]
