Answer :
Answer:
A) [tex]a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}[/tex]
B) [tex]a_{0} = a_{1} = 0[/tex]
C) for n = 2
[tex]a_{2}[/tex] = 1
for n = 3
[tex]a_{3}[/tex] = 3
for n = 4
[tex]a_{4}[/tex] = 8
for n = 5
[tex]a_{5}[/tex] = 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be : [tex]2^{n-2}[/tex] strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
[tex]a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}[/tex]
b ) The initial conditions
The initial conditions are : [tex]a_{0} = a_{1} = 0[/tex]
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
[tex]a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}[/tex]
for n = 2
[tex]a_{2}[/tex] = 1
for n = 3
[tex]a_{3}[/tex] = 3
for n = 4
[tex]a_{4}[/tex] = 8
for n = 5
[tex]a_{5}[/tex] = 19