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(A) For an $n$-digit integer with no consecutive $0$s:If it ends in $1$,the previous digit can be $0$ or $1$. Thus,$b_n = b_{n-1} + c_{n-1} = a_{n-1}$

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$(A, B)$

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(A) For an $n$-digit integer with no consecutive $0$s:If it ends in $1$,the previous digit can be $0$ or $1$. Thus,$b_n = b_{n-1} + c_{n-1} = a_{n-1}$.If it ends in $0$,the previous digit must be $1$. Thus,$c_n = b_{n-1}$.Since $a_n = b_n + c_n$,we have $a_n = a_{n-1} + a_{n-2}$.$1.$ For $(A)$,$a_{17} = a_{16} + a_{15}$ is true by the recurrence relation.For $(B)$,$c_{17} = c_{16} + c_{15}$ is true,so $c_{17} 
eq c_{16} + c_{15}$ is false.For $(C)$,$b_{17} = b_{16} + c_{16}$ is true,so $b_{17} 
eq b_{16} + c_{16}$ is false.For $(D)$,$a_{17} = b_{17} + c_{17} = (b_{16} + c_{16}) + c_{17} = a_{16} + c_{17}$. Since $a_{16} = b_{16} + c_{16}$,this does not simplify to $a_{17} = c_{17} + b_{16}$.Thus,only $(A)$ is correct.$2.$ We have $b_n = a_{n-1}$.$a_1 = 1$ $(1)$$a_2 = 2$ $(10, 11)$$a_3 = 3$ $(101, 110, 111)$$a_4 = 5$ $(1010, 1011, 1101, 1110, 1111)$$a_5 = 8$Therefore,$b_6 = a_5 = 8$.Correct option is $(B)$.

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