Question

Let a_n denote the number of all n-digit positive integers formed by the digit 0,1 or both such that no consecutive digit 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.
I. Which of the following is correct of:
(a) a₁₇ = a₁₆ + a₁₅ (b) c₁₇ is not = c₁₆ + c₁₅
(c) b₁₇ is not = b₁₆ + c₁₆ (d) a₁₇ = c₁₇ + b₁₆
II. The value of b₆ is :
(a) 7 (b) 8 (c) 9 (d) 11

Answer 1

$$\begin{gathered} \mathrm{Note}\mathrm{that}{\mathrm{a}}_1=1\mathrm{and}{\mathrm{a}}_2=2 \\ \mathrm{For}\mathrm{n}\ge 3,\mathrm{if}\mathrm{the}\mathrm{last}\mathrm{digit}\mathrm{is}1,\mathrm{then}\mathrm{first}\left(\mathrm{n}-1\right)\mathrm{digit}\mathrm{can}\mathrm{be}\mathrm{choosen}\mathrm{in}{\mathrm{a}}_{\mathrm{n}-1} \\ \mathrm{ways}\mathrm{and}\mathrm{if}\mathrm{the}\mathrm{last}\mathrm{digit}\mathrm{is}0,\mathrm{then}\left(\mathrm{n}-1\right)\mathrm{th}\mathrm{digit}\mathrm{is}1\mathrm{and}\mathrm{the}\mathrm{first}\left(\mathrm{n}-1\right) \\ \mathrm{digits}\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}{\mathrm{a}}_{\mathrm{n}-2}\mathrm{ways},\mathrm{Thus} \\ {\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}+{\mathrm{a}}_{\mathrm{n}-2}\forall \mathrm{n}\ge 3......................\left(1\right) \\ \mathrm{Also}\mathrm{note}\mathrm{that}{\mathrm{b}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}\forall \mathrm{n}\ge 3 \\ \mathrm{and}{\mathrm{c}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-2}\forall \mathrm{n}\ge 3 \\ \mathrm{Now}\mathrm{we}\mathrm{have}, \\ {\mathrm{b}}_6={\mathrm{a}}_5={\mathrm{a}}_4+{\mathrm{a}}_3 \\ ={\mathrm{a}}_3+{\mathrm{a}}_2+{\mathrm{a}}_3 \\ =2\left({\mathrm{a}}_2+{\mathrm{a}}_1\right)+{\mathrm{a}}_2 \\ =3\left(2\right)+2\left(1\right) \\ =8 \\ \mathrm{And}\mathrm{putting}\mathrm{n}=17\mathrm{in}\mathrm{equation}1\mathrm{we}\mathrm{get}, \\ {\mathrm{a}}_{17}={\mathrm{a}}_{16}+{\mathrm{a}}_{15} \end{gathered}$$