Question

Let $a_n$ denote the number of all $n$ digit positive integers formed by the digits $0,1,$ of 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

• A. $7$
• B. $8$
• C. $9$
• D. $11$

Answer 1

The correct option is B

$8$

Explanation for the correct option:

Finding the value of $b_6$:

Given $a_n$ denote the number of all $n$ digit positive integers formed by the digits $0,1,$

Therefor,

For $a_1$total number $=1$ since we have only '$1$' as a positive integer formed by the digits $0,1,$

For $a_2$total numbers $=2$ As we have numbers $\left\{10,11\right\}$

For $a_3$total numbers $=3$ As we have numbers $\left\{101,110,111\right\}$

According to given information we will not take $100$ in above set because two consecutive digits are zero.

For $a_4$total numbers $=5$ As we have numbers $\left\{1010,1011,1101,1111,1110\right\}$

Observing $a_1$, $a_2$, $a_3$,$a_4$ we get a relation

$a_n=a_{n-1}+a_{n-2}$

Also, given $b_n=$ the number of such $n$ digit integers ending with digit $1$

Similarly

$\begin{array}{rcl}{b}_1& =& \left\{1\right\}\\ b_2& =& \left\{11\right\}\\ b_3& =& \left\{101,111\right\}\\ b_4& =& \left\{1011,1101,1111\right\}\end{array}$

Similarly we get the relation,

$b_n=b_{n-1}+b_{n-2},\forall n>3$

Substituting $n=6$

$\begin{array}{rcl}{b}_6& =& b_{6-1}+b_{6-2}\\ & =& b_5+b_4\\ & =& \left(b_4+b_3\right)+b_4\mathbf{[}\mathbf{}\mathbf{\text{Again by this relation}}\mathbf{]}\\ & =& 2\times \left(\text{number of element in}{b}_4\right)\mathbf{}\mathbf{+}\left(\text{number of element in}{b}_3\right)\\ & =& 2\times 3+2\\ b_6& =& 8\end{array}$

Hence, option (B) is the correct answer.