Question

a_n = numbers of all n digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are zero.

  $\Rightarrow$ the number will end with 0 or 1

Case-I:  If the number ends with 0, then 

(n – 1)^th digit should be 1

Hence number of such numbers will be b_{n – 1}.

Case-II: If the number ends with 1, then

First (n – 1) digits should be (n – 1) digit positive integers formed by

the digits 0, 1 or both such that no consecutive digits in them are zero.

Hence number of such numbers will be a_{n – 1}.

 $\therefore$ an = a_{n – 1} + b_{n – 1}                                .......(1)

but   b_n is the number of such numbers which will end with 1

 $\therefore$ first (n – 1) digits will be (n – 1) digit positive integers formed by

the digits 0, 1 or both such that no consecutive digits in them are zero.

Hence b_n = a_{n – 1}

From (1), we get

a_n = a_{n – 1} + a_{n – 2} and for n = 17, a₁₇ = a₁₆ + a₁₅                           

c_n is the number of such numbers ending with 0.

 $\therefore$   (n – 1)^th digit should be  1.

first (n – 1) digits will be (n – 1) digit positive integers formed by

the digits 0, 1 or both such that no consecutive digits in them are zero

which will end with 1.

 $\therefore$   c_n = b_{n – 1}.

b₆ = a₅ = a₄ + a₃ = 2a₃ + a₂ = 3a₂ + 2a1

a₂ = 2                (10 and 11)

 a₁ = 1                (1)

  $\therefore$ b₆ = 8