The goal is to find a recurrence relation for the number of bit strings of length $n$ that contain the substring "01." Let's break down the explanation step by step:

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Defining the Problem

Let $a_n$ represent the number of bit strings of length $n$ that contain the substring "01."

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Case 1: Strings Starting with 1

If the bit string starts with $1$, the remaining $n-1$ bits must form a string of length $n-1$ that contains "01." By definition, there are $a_{n-1}$ such bit strings.

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Case 2: Strings Starting with $k$ Consecutive Zeros Followed by a 1

If the string starts with $k$ consecutive zeros (where $1 \leq k \leq n-1$) and is then followed by a $1$, we only need to focus on the remaining $n-k-1$ bits. These remaining bits can be any combination of $0$s and $1$s, meaning there are $2^{n-k-1}$ possible strings for each $k$.

Summing Over All $k$:

For all $k$ from $1$ to $n-1$, the total number of strings is: $\sum_{k=1}^{n-1} 2^{n-k-1}$

This simplifies to a geometric series: $2^{n-2} + 2^{n-3} + \cdots + 2^0 = 2^{n-1} - 1$

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Combining Both Cases

Adding both contributions gives us:

1. From Case 1: $a_{n-1}$
2. From Case 2: $2^{n-1} - 1$

Thus, the recurrence relation is: $a_n = a_{n-1} + (2^{n-1} - 1)$

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Base Case

For $n = 2$, we can calculate $a_2$ explicitly:

• The bit strings of length $2$ are: $00, 01, 10, 11$.
• Only $01$ contains the substring "01." Thus, $a_2 = 1$.

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Summary

The recurrence relation: $a_n = a_{n-1} + (2^{n-1} - 1), \quad \text{valid for } n \geq 2$ captures the process of building bit strings of length $n$ that contain "01," either by extending shorter strings that already contain "01" or by placing "01" in specific configurations in the new string.

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Would you like additional details or examples to clarify? Here are 5 related questions for exploration:

1. How do we explicitly compute the first few values of $a_n$?
2. Can we derive a closed-form solution for $a_n$ using this recurrence?
3. How does the geometric series simplify to $2^{n-1} - 1$?
4. What happens if we look for bit strings that contain "10" instead of "01"?
5. Can this recurrence relation be modified to count bit strings containing "01" and "10"?

Tip: Recurrence relations often help to break down complex counting problems by connecting smaller cases to larger ones.