Let denote the number of all -digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let = the number of such -digit integers ending with digit 1 and = the number of such -digit integers ending with dig

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. [2012]

Q. The value of $b_{6}$ is

Smart Achievers · Accepted Answer

(2)

$∵$ $n$

$and b_{n} = number of such n -digit integers ending with 1$

$and c_{n} = number of such n -digit integers ending with 0$

$Clearly, a_{n} = b_{n} + c_{n}$ $(∵ a_{n} can end with 0 or 1)$

$Also b_{n} = a_{n - 1}$

$and c_{n} = a_{n - 2}$ $[∵ if last digit is 0, second last has to be 1]$

$∴$ $a_{n} = a_{n - 1} + a_{n - 2}, n \geq 3$

$Also a_{1} = 1, a_{2} = 2$

$Now by this recurring formula, we get$

$a_{3} = a_{2} + a_{1} = 3$

$a_{4} = a_{3} + a_{2} = 3 + 2 = 5$

$a_{5} = a_{4} + a_{3} = 5 + 3 = 8$

$Also b_{6} = a_{5} = 8$

Solution

1 7

2 8

3 9

4 11

Want Us to Email you About Special Offers & Updates?