JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Sequences and Series

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.

The value of

b_{6}

Answer: (B)

Solution

We have the recurrence relations:

$a_{n} = b_{n} + c_{n}$ (total = ending in 1 + ending in 0)
$b_{n} = a_{n - 1}$ (to end in 1, the previous $n - 1$ digits can be any valid configuration)
$c_{n} = b_{n - 1}$ (to end in 0, the previous digit must be 1)

From these:

a_{n} = a_{n - 1} + b_{n - 1} = a_{n - 1} + a_{n - 2}

(Fibonacci-like).

Starting values:

a_{1} = 1

(only "1"),

a_{2} = 2

("10", "11").

Computing:

a_{3} = 3

a_{4} = 5

a_{5} = 8

a_{6} = 13

.

Since

b_{6} = a_{5} = 8

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option