Let \(a_n\) represent the number of bit strings of length \(n\) containing two…

Key idea: partition all n-bit strings by their first one or two bits.

• If the first two bits are 11, the remaining n-2 bits can be anything, yielding 2^{n-2} strings that contain two consecutive 1s.

• If the first bit is 0, then the last n-1 bits must contain two consecutive 1s, contributing a_{n-1} strings.

• If the first two bits are 10, then the last n-2 bits must contain two consecutive 1s, contributing a_{n-2} strings.

These three cases are disjoint and cover all n-bit strings, so

a_n = a_{n-1} + a_{n-2} + 2^{n-2}.

Base cases:

• a_0 = 0 (empty string contains no 11).

• a_1 = 0 (strings of length 1 cannot contain 11).

• a_2 = 1 (only the string 11 contains consecutive 1s).

Thus the recurrence and base cases fully determine a_n.