Find Number of Valid Binary Integers Ending in One

To find the value of $b_6$, we first analyze the rules for forming the $n$-digit integers using the digits $0$ and $1$:

1. Since the numbers are positive $n$-digit integers, the first digit must always be $1$ (it cannot be $0$).
2. No two consecutive digits can be $0$.
3. $b_n$ is the number of such $n$-digit integers ending in $1$.
4. $c_n$ is the number of such $n$-digit integers ending in $0$.
5. $a_n = b_n + c_n$ is the total number of such $n$-digit integers.

Recurrence Relations:

• An $n$-digit valid integer ending in $1$ can be formed by appending $1$ to any valid $(n-1)$-digit integer (ending in either $1$ or $0$). Hence: $b_n = b_{n-1} + c_{n-1} = a_{n-1} \quad \text{for } n \ge 2$

• An $n$-digit valid integer ending in $0$ can only be formed by appending $0$ to a valid $(n-1)$-digit integer that ends in $1$ (to avoid consecutive zeros). Hence: $c_n = b_{n-1} \quad \text{for } n \ge 2$

• The total number of valid $n$-digit integers satisfies: $a_n = b_n + c_n = a_{n-1} + a_{n-2} \quad \text{for } n \ge 3$

Computing Values Step-by-Step:

• For $n = 1$:

  • Valid integers: $\{1\}$
  • $b_1 = 1$, $c_1 = 0 \implies a_1 = 1$

• For $n = 2$:

  • $b_2 = a_1 = 1$ (integer: $11$)
  • $c_2 = b_1 = 1$ (integer: $10$)
  • $a_2 = b_2 + c_2 = 2$

• For $n = 3$:

  • $b_3 = a_2 = 2$ (integers: $101, 111$)
  • $c_3 = b_2 = 1$ (integer: $110$)
  • $a_3 = b_3 + c_3 = 3$

• For $n = 4$:

  • $b_4 = a_3 = 3$
  • $c_4 = b_3 = 2$
  • $a_4 = b_4 + c_4 = 5$

• For $n = 5$:

  • $b_5 = a_4 = 5$
  • $c_5 = b_4 = 3$
  • $a_5 = b_5 + c_5 = 8$

• For $n = 6$:

  • $b_6 = a_5 = 8$

Thus, the value of $b_6$ is $8$.

Correct Option: (B)