Question

Let $a_n$ 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 $0$.
Which of the following is correct $?$

• A. $a_{17}=a_{16}+a_{15}$
• B. $c_{17}\ne c_{16}+c_{15}$
• C. $b_{17}\ne b_{16}+c_{16}$
• D. $a_{17}=c_{17}+b_{16}$

Answer 1

The correct option is A $a_{17}=a_{16}+a_{15}$

$a_n:n$ digit positive terms by $0,1$ in which no two consecutive $0^{\prime }s$ exist

Case $1:1$ _ _ _ _ _ $1$ (end with $1$)
$a_{n-1}=b_n$

Case $2:1$ _ _ _ _ _ $0$ (end with $0$)
$a_{n-2}=c_n$

$\therefore a_n=a_{n-1}+a_{n-2}\Rightarrow a_n=b_n+c_n$
$\Rightarrow a_{17}=a_{16}+a_{15}$

$c_{17}=c_{16}+c_{15}$
$a_{17}=b_{16}+c_{17}$

$\therefore$ option $\left(a\right)$ is correct and $\left(b\right),\left(c\right)$ and $\left(d\right)$ are false.