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 ${c_n} =$ Then number of such n-digit integers with digit $0.$ The value of ${b_6}$ is
A) $7$
B) $8$
C) $9$
D) $11$

Answer 1

Here the term ${b_n}$ is representing n-digit positive integers formed by digits $0,\;1,$ or both such that no consecutive digits in them are $0$and also ending with $1.$ To find the value of ${b_6}$, find values of some starting terms of ${a_n}$ and then find relation between the terms, similarly find some ${b_n}$ terms and relation between them. And use that relation to get required value.

Complete step-by-step solution:
To find the value of ${b_6}$ we will find some terms of ${a_n}$ and then try to find relation between the terms, as follows

$${a_1} = 1 \\\ {a_2} = 10,\,11 \\\ {a_3} = 101,\;111,\,110 \\\ {a_4} = 1010,\;1011,\;1101,\;1110,\;1111 \\\$$

Now we can see that,

$${a_1} + {a_2} = {a_3} \\\ {a_2} + {a_3} = {a_4} \\\$$

Similarly we can write that,

$${a_3} + {a_4} = {a_5} \\\ {a_4} + {a_5} = {a_6} \\\$$

Now, in ${a_1}$ we can see that there is no ${c_1}$ but there exist a ${b_1}$ that is $1$
And in ${a_2}$ there is one ${b_1}$ that is $11$
And similarly we can write that,
${b_3} = {b_1} + {b_2} = 1 + 1 = 2 \\\ {b_4} = {b_2} + {b_3} = 1 + 2 = 3 \\\$
Going on further, we will get
${b_5} = {b_3} + {b_4} = 2 + 3 = 5 \\\ {b_6} = {b_4} + {b_5} = 3 + 5 = 8 \\\$

So we get the required value of ${b_6} = 8$

Note: This type of problems does not have a particular way to be solved but they can be solved by understanding the algorithm in them. So when you tackle this type of question again then your first step should be to write a few terms, so that you will get a physical appearance about the algorithm and then you can solve it more clearly.