Value of $b_{5}\;,b_{6}\;\;b_{7}.$ If $b_{n}$ Rep. no. of $n$ digit integer ending with $1$ 
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 $0$ and $b_{n}$ denote number of such
$n$ digit integers ending with digit $1$ and $c_{n}$ denote number of such
  $n$ digit integers
ending with digit $0$.
$(i)\;\;\;$ Then calculate the values $b_{5}\;,b_{6}\; ,b_{7}$
$(ii)\;\; $Also, find which of the following is right?

$\bf{Options::}$ 

$(a)\;\; a_{17}=a_{16}+a_{15}\;\;\;\;\;(b)\;\; c_{17}\neq c_{16}+c_{15}\;\;\;\;\;(c)\;\;b_{17}\neq b_{16}+c_{16}\;\;\;\;\; (d)\;\ a_{17}=c_{17}+b_{16}$ 

$\bf{My\; Try::}$ Here  Given $a_{n}$ Represent all $n$ Digit positive integer ending with $0,1$
and $b_{n}$ Represent such all $n$ Digit positive Integer ending with $1$
and $c_{n}$ Represent such all $n$ Digit positive Integer ending with $0$
So we get  $a_{n} = b_{n}+c_{n}.$
and $a_{1} = 1(1)\;\;,a_{2}=2(10,11)\;\;,a_{3}=3(110,101,111)\;\;,a_{4}=5(1111,1110,1101,1011,1010)$
and $b_{1}=1(1)\;\;,b_{2}=1(11)\;\;,b_{3}=2(101,111)\;\;,b_{4}=3(1111,1101,1011)$
Similarly $c_{1}=0\;\;,c_{2}=1(10)\;\;,c_{3} =1(110)\;\;,c_{4}=2(1110,1010)$
Now I did not understand How can I solve after that.
Help Required.
Thanks.
 A: By definition $a_n = b_n + c_n$.
If we have a correct number of length $n+1$ ending in $1$, then removing the final $1$ will always result in a correct number of length $n$ (of which there are $a_n$). And if we have a correct number of length $n$, we can always append a 1 to it, and get a correct number of lenght $n+1$, ending in $1$ (of which there are $b_{n+1}$), because the $1$ can never cause the 2 consecutive $0$ condition to fail.
So $b_{n+1} = a_n$ for all $n$. (This also holds for your small number examples, as you can see).
If we have a correct number of length $n+1$ ($n \ge 2$) ending in $0$, then it must end in $10$ as $00$ would have been illegal. Removing the final $10$ leaves us with a correct number of length $n-1$, of which there are $a_{n-1}$ by definition. Conversely again, having a correct number of length $n-1$ we can always add $10$ to it, which will never spoil the correctness, and get a valid number of length $n+1$ ending in $0$. 
So $c_{n+1} = a_{n-1}$, as also can be seen from your examples.
So $a_{n+1} = b_{n+1} + c_{n+1} = a_n + a_{n-1}$ for all $n \ge 2$. I.e. we have a Fibonacci sequence.
