Let n denote the number of all n-digit positive integers formed by the digits `0, 1` or both such that no consecutive digits in them are 0. Let `b_n` = the number of such n-digit integers ending with digit 1 and `c_n`= the number of such n-digit integers ending with digit 0. The value of `b_6`, is

Options

A. 7
B. 8
C. 9
D. 11

Answer

B

Explanation

`because a_(n)=`number of all n-digit positive integers formed by the digits 0,1 or both such that no consecutive digits in them are zero.
and `b_(n)=`number of such n-digit integers ending with 1
`c_(n)=`number of such n-digit integers ending with 0.
clearly `a_(n)=b_(n)+c_(n)" "[becausea_(n)" can end with 0 or 1"]`
also, `b_(n)=a_(n-1) and c_(n)=a_(n-2)` [`because` if last digit is 0, second last has to 1]
`therefore` We get `a_(n)=a_(n-1)+a_(n-2),n ge 3`
also, `a_(1)=1,a_(2)=2`
by the recurring formula `a_(3)=a_(2)+a_(1)=3`
`a_(4)=a_(3)+a_(2)=3+2=5`
`a_(5)=a_(4)+a_(3)=5+3=8`
also, `b_(6)=a_(5)=8`