To explain the answer systematically, let's break it down step by step:

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Understanding the Problem

We are tasked to find a recurrence relation for the number of bit strings of length $n$ that contain the substring "01". A recurrence relation expresses the value of $a_n$ (the solution for length $n$) in terms of $a_{n-1}$ (the solution for length $n-1$) and possibly other terms.

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Analyzing the Solution

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

1. Case 1: Start with "1"

   • If a bit string of length $n$ starts with "1," then the remaining $n-1$ bits must contain the substring "01."
   • The number of such bit strings is $a_{n-1}$ because we can take any valid bit string of length $n-1$ containing "01" and prepend a "1" to it.

2. Case 2: Start with $k$ "0"s and then "1"

   • Consider a bit string that starts with $k$ zeros (where $1 \leq k \leq n-1$) followed by a "1."

   • After the initial segment of $k$ zeros and "1," the remaining $n-k-1$ bits are free to be anything (since no restrictions are imposed on them).

   • For each $k$, there are $2^{n-k-1}$ such strings (since there are $2^{n-k-1}$ possible combinations for the last $n-k-1$ bits).

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Summing Over All $k$

For $k$ ranging from 1 to $n-1$, the total number of strings of this type is: $\sum_{k=1}^{n-1} 2^{n-k-1}.$ This is a geometric series: $2^0 + 2^1 + 2^2 + \dots + 2^{n-2}.$ The sum of this series is: $2^{n-1} - 1.$

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Combining the Two Cases

The total number of bit strings of length $n$ containing "01" is the sum of the two cases:

1. Bit strings starting with "1" and containing "01": $a_{n-1}$.
2. Bit strings starting with $k$ "0"s followed by "1": $2^{n-1} - 1$.

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

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Validity of the Relation

• This recurrence is valid for $n \geq 2$, since for $n = 1$, it's impossible to have the substring "01."

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Conclusion

The recurrence relation for the number of bit strings of length $n$ containing "01" is: $a_n = a_{n-1} + (2^{n-1} - 1),$ where $n \geq 2$.

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Let me know if you need further clarification or have any related questions!

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Five Related Questions

1. What is the base case $a_1$ for this recurrence relation?
2. How would you compute $a_3$ explicitly using this recurrence relation?
3. How can you verify that the geometric series formula $2^{n-1} - 1$ is correct for the sum?
4. What is the closed-form solution for this recurrence relation, if any?
5. How does this recurrence relation compare to one for bit strings containing "00" instead of "01"?

Tip

For solving recurrence relations involving bit strings, carefully consider both restrictions and freedoms in constructing the strings.