Let $a_n$ represent the number of bit strings of length $n$ containing two consecutive 1s. What is the recurrence relation for $a_n$?

To construct a bit string of length $n$ containing two consecutive 1s, we can condition on how the string ends:

1. The string ends in $0$: The remaining prefix of length $n-1$ must contain two consecutive 1s. The number of such strings is $a_{n-1}$.

2. The string ends in $01$: The remaining prefix of length $n-2$ must contain two consecutive 1s. The number of such strings is $a_{n-2}$.

3. The string ends in $11$: The condition of having two consecutive 1s is already satisfied by the suffix 11. Hence, the preceding $n-2$ bits can be any arbitrary bit string. The number of such strings is $2^{n-2}$.

Summing these mutually exclusive and exhaustive cases, we get the recurrence relation:

Hence, option A is the correct answer.