Discussion Forum



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. 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.

Which of the following is correct?


A) $a_{17}=a_{16}+a_{15}$

B) $a_{17} \neq a_{16}+a_{15}$

C) $b_{17} \neq b_{16}+c_{16}$

D) $a_{17}=c_{17}+b_{16}$

Answer:

Option A

Explanation:

 Since $a_{n}$ be the n-digit positive integer formed by the digits 0,1 or both such that no two  consecutive digits are zero.

 $b_{n}$=numbers which are ending with 1

$C_{n}$= number which are ending with 0

$\therefore$  $a_{n}=b_{n}+c_{n}$

 i.e, $a_{n}$=1(0 or 1)...............(0 or 1)

$b_{n}=$1 (0 or 1) ........1   

                                            nth place

 $c_{n}$=1(0 or 1) ......0

As   $a_{n}=b_{n}+c_{n}$

 i.e,   $a_{n}=1$.................(1 or 0).

18112021510_k9.PNG

$\Rightarrow$  $a_{n}=a_{n-1}+a_{n-2}$

 $\therefore$    $a_{17}=a_{16}+a_{15}$