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. Which of the following is correct

Q. Which of the following is correct?

Smart Achievers · Accepted Answer

(1)

$By recurring formula, a_{17} = a_{16} + a_{15} is correct.$

$n$

$\Rightarrow a_{15} \neq a_{14} + a_{13}$ $(∵ c_{n} = a_{n - 2})$ $∴$ $Incorrect$

$Similarly, other parts are also incorrect.$

Solution

1 $a_{17} = a_{16} + a_{15}$

2 $c_{17} \neq c_{16} + c_{15}$

3 $b_{17} \neq b_{16} + c_{16}$

4 $a_{17} = c_{17} + b_{16}$

Ans.
(1)

$By recurring formula, a_{17} = a_{16} + a_{15} is correct.$

$Also c_{17} \neq c_{16} + c_{15}$

$\Rightarrow a_{15} \neq a_{14} + a_{13}$ $(∵ c_{n} = a_{n - 2})$ $∴$ $Incorrect$

$Similarly, other parts are also incorrect.$

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Solution

1 a17=a16+a15

2 c17≠c16+c15

3 b17≠b16+c16

4 a17=c17+b16