How many length n binary numbers have no consecutive zeroes ?Why we get a Fibonacci pattern? [duplicate]

Consider the length $n$ binary strings with no repeated zeros (BSWNRZ).

BSWNRZ must end in a $0$ or a $1$. The number of BSWNRZ of length $n$ that end in a $0$ is equal to the number of BSWNRZ of length $n-2$ (with $10$ appended). The number of BSWNRZ of length $n$ that end in a $1$ is equal to the number of BSWNRZ of length $n-1$ (with $1$ appended).

That is, any length $n$ BSWNRZ falls into one of the following two cases

$$ \overbrace{?????????????????}^{\text{any length }n-2\text{ BSWNRZ}}\color{#00A000}{1}\color{#C00000}{0}\\ \underbrace{??????????????????}_{\text{any length }n-1\text{ BSWNRZ}}\color{#C00000}{1} $$

Therefore, the number of BSWNRZ of length $n$ is equal to the number of BSWNRZ of length $n-1$ plus the number of BSWNRZ of length $n-2$. Thus, the Fibonacci pattern. $$ F_n=F_{n-1}+F_{n-2} $$