Final answer:
To create a recurrence relation for the number of bit strings, we can use the fact that each n-bit string can generate two (n+1)-bit strings. This leads to the relation an+1 = 2 × an, with the base case a0 = 1.
Step-by-step explanation:
To find a recurrence relation for the number of bit strings, let's denote the number of n-bit strings without restrictions as an. Each bit in a string can either be 0 or 1, which means that for any n-length bit string, you can create two (n+1)-length bit strings by appending a 0 or a 1 to the original string. Thus, the recurrence relation for the unrestricted number of bit strings of length (n+1) is based on the number of n-length strings.
The recurrence relation can be expressed as:
an+1 = 2 × an
This relation reflects that for every bit string of length n, there are exactly two bit strings of length (n+1), one with a 0 appended and one with a 1 appended.
To start the sequence, we need a base case. For example, the number of 0-length strings (a string with no bits) is 1, which serves as our base case: a0 = 1.
Thus, for a given string of length 1, the number would be a1 = 2, and so on.