For each integer n ≥ 0 . let a k be the number of bit strings of length n that do not contain the pattern 101.

a. Show that a k = a k − 1 + a k − 3 + a l − 4 + ⋯ + a 0 + 2 , for every integer k ≥ 3 .
b. Use the result of part (a) to show that if k ≥ 3 , then a k = 2 a k − 1 + a k − 2 + a k − 3 .

(a)

To show that ak=ak−1+ak−3+ak−4+.........+a0+2 ,for all integers k≥3.

Given information:

For each integer n≥0, let ak ,be the number of bit strings of length n that do not contain the pattern 101.

Concept used:

If the string starts with 0, it is followed by any string of k−1 bits that does not contain the pattern 101.

Calculation:

ak= The number of bit strings of length k that do not contain the pattern 101.

Let k be an integer with k≥3.

Now, any string of length k that does not contain the pattern 101 starts either with 0 or 1.

If the string starts with 0, it is followed by any string of k−1 bits that does not contain the pattern

101.

Thus, there are ak−1 strings of this category

If the string starts with 1, and if the first three bits are 100, then it is followed by any string of k−3 bits that does not contain the pattern 101.

Thus, there are ak−3 strings of this category.

If the string starts with 1 and if the first four bits are 1100, then it is followed by followed by any strings of k−4 bits that does not contain the pattern 101.

Thus, there are ak−4 strings of this category.

If the strings starts with 1, and if the first five bits are 11100, then it is followed by any string of k−5 bits that does not contain the pattern 101

(b)

To show that if k≥3, then

ak=2ak−1−ak−2+ak−3 using part (a).