@divinedolphin This is such a clever comment! I really like it! You are absolutely correct that another possible answer is
$$ c_n = a_{n-2} $$
It just doesn't happen to be one of the answer choices.
Let's take a look at the given solution provided first. Our sequences are: \(a_{n},\) denoting sequences ending with a black bead, \(b_n,\) denoting sequences ending with only one white bead, and \(c_n,\) denoting sequences ending with two white beads.
In order to make a \(c_n\) sequence ending with two white beads, we can start with a \(b_{n-1}\) sequence of length \( n-1\) ending with one white bead. We cannot start with an \(a_{n-1}\) sequence, since that would give us \( \textcolor{blue}{{BW}}, \) and we cannot start with a \(c_{n-1}\) sequence, since that would give us \( \textcolor{red}{WWW},\) which isn't allowed.
The number of \(\textcolor{blue}{WW}\) sequences is exactly the number of \(\textcolor{blue}{BW}\) sequences of length \( n-1.\)
$$ \boxed{ c_n = b_{n-1}} $$
This was the given answer to the mini-question.
Now we can iterate one step farther back. Let's think; how do we get a \(b_{n-1}\) sequence?
If we start with a \(b_{n-2}\) sequence and add a white bead, we'll get two white beads at the end, which isn't a \(b_{n-1}\) sequence. If we start with a \(c_{n-2}\) sequence and add a white bead, we'll get three beads at the end, which isn't allowed. The only way to get a \(b_{n-1}\) sequence is to start with a \(a_{n-2}\) sequence, which ends in black, and to add a white bead at the end.
$$ b_{n-1} = a_{n-2} $$
And from before, we know
$$\begin{aligned}
c_{n} &= b_{n-1} \\
&= a_{n-1} \\
\end{aligned}
$$
So you are correct in thinking that \(a_{n-2}\) could be another possible answer choice.