Let `a_(n)` denote the number of all n-digit numbers formed by the digits 0,1 or both such that no consecutive digits in them are 0. Let `b_(n)` be the number of such n-digit integers ending with digit 1 and let `c_(n)` be the number of such n-digit integers ending with digit 0. Which of the following is correct ?

Options

A. `a_(17)=a_(16)+a_(15)`
B. `c_(17)nec_(16)+c_(15)`
C. `b_(17)neb_(16)+c_(16)`
D. `a_(17)=c_(17)+b_(16)`

Explanation

To solve the problem, we need to analyze the sequences defined by \( a_n \), \( b_n \), and \( c_n \).

### Step 1: Define the sequences
- Let \( a_n \) be the total number of \( n \)-digit numbers formed by the digits 0 and 1 such that no two consecutive digits are 0.
- Let \( b_n \) be the number of such \( n \)-digit integers ending with the digit 1.
- Let \( c_n \) be the number of such \( n \)-digit integers ending with the digit 0.

### Step 2: Establish relationships
1. **For \( a_n \)**:
- An \( n \)-digit number can either end with 1 or 0.
- If it ends with 1, the first \( n-1 \) digits can be any valid \( (n-1) \)-digit number, which is \( a_{n-1} \).
- If it ends with 0, the \( (n-1) \)-th digit must be 1 (to avoid two consecutive 0s), and the first \( n-2 \) digits can be any valid \( (n-2) \)-digit number, which is \( b_{n-1} \).
- Therefore, we have:
\[ a_n = b_n + c_n \]

2. **For \( b_n \)**:
- An \( n \)-digit number ending with 1 can be formed by appending 1 to any valid \( (n-1) \)-digit number, which can end with either 0 or 1. Thus:
\[ b_n = a_{n-1} \]

3. **For \( c_n \)**:
- An \( n \)-digit number ending with 0 must have the \( (n-1) \)-th digit as 1. Thus, it can be formed by appending 0 to any valid \( (n-1) \)-digit number that ends with 1. Therefore:
\[ c_n = b_{n-1} \]

### Step 3: Recurrence relations
From the relationships established:
- We can express \( a_n \) in terms of \( a_{n-1} \) and \( a_{n-2} \):
\[ a_n = a_{n-1} + b_{n-1} \]
Substituting \( b_{n-1} = a_{n-2} \):
\[ a_n = a_{n-1} + a_{n-2} \]

- Similarly, we can express \( b_n \) and \( c_n \):
\[ b_n = a_{n-1} \]
\[ c_n = b_{n-1} = a_{n-2} \]

### Step 4: Conclusion
From the above, we can summarize:
- \( a_n = a_{n-1} + a_{n-2} \)
- \( b_n = a_{n-1} \)
- \( c_n = a_{n-2} \)