In the non-decreasing sequence of odd integers `(a_(1),a_(2),a_(3),....)={1,3,3,3,5,5,5,5,5....}` each positive odd integer k appears k times. It is a fact that there are integers b,c and d such that for all positive integer `n,a_(n)=b[sqrt(n+c)]+d` (where [.] denotes greatest integer function). The possible vaue of b+c+d is

Options

A. (a)0
B. (b)1
C. (c)2
D. (d)4

Answer

To solve the problem, we need to find integers \( b \), \( c \), and \( d \) such that for all positive integers \( n \), the sequence \( a_n \) can be expressed as:

\[ a_n = b \cdot \lfloor \sqrt{n + c} \rfloor + d \]

where \( \lfloor . \rfloor \) denotes the greatest integer function.

### Step 1: Understanding the Sequence
The sequence given is \( 1, 3, 3, 3, 5, 5, 5, 5, 5, \ldots \). Each positive odd integer \( k \) appears \( k \) times.

### Step 2: Finding the Pattern
We can observe that:
- The number 1 appears 1 time.
- The number 3 appears 3 times.
- The number 5 appears 5 times.

The total number of terms up to the odd integer \( k \) is given by the sum of the first \( k \) odd integers, which is \( 1 + 3 + 5 + \ldots + k = \frac{k(k + 1)}{2} \).

### Step 3: Determining the Range for \( n \)
For a given odd integer \( k \):
- The first occurrence of \( k \) is at position \( \frac{(k-1)k}{2} + 1 \).
- The last occurrence of \( k \) is at position \( \frac{k(k + 1)}{2} \).

### Step 4: Relating \( n \) to \( k \)
To find \( k \) in terms of \( n \), we can solve:
\[ \frac{(k-1)k}{2} < n \leq \frac{k(k + 1)}{2} \]

This implies that \( k \) can be approximated as:
\[ k \approx \sqrt{2n} \]

### Step 5: Setting Up the Equation
We can assume \( k \) is approximately \( \lfloor \sqrt{2n} \rfloor \) for large \( n \). Therefore, we can write:
\[ a_n = b \cdot \lfloor \sqrt{n + c} \rfloor + d \]

### Step 6: Finding \( b \), \( c \), and \( d \)
From our observations:
- Since \( a_n \) should yield odd integers, we can set \( b = 2 \) (to account for the odd integers).
- We also need to find \( c \) and \( d \).

Using the first term \( a_1 = 1 \):
\[ 1 = 2 \cdot \lfloor \sqrt{1 + c} \rfloor + d \]

### Step 7: Solving for \( c \) and \( d \)
Assuming \( d = 1 \):
\[ 1 = 2 \cdot \lfloor \sqrt{1 + c} \rfloor + 1 \]
\[ 0 = 2 \cdot \lfloor \sqrt{1 + c} \rfloor \]
This implies \( \lfloor \sqrt{1 + c} \rfloor = 0 \), which leads to \( 1 + c < 1 \) or \( c < 0 \).

The only integer satisfying this is \( c = -1 \).

### Step 8: Final Values
Thus, we have:
- \( b = 2 \)
- \( c = -1 \)
- \( d = 1 \)

### Step 9: Calculating \( b + c + d \)
\[ b + c + d = 2 + (-1) + 1 = 2 \]

### Conclusion
The possible value of \( b + c + d \) is \( 2 \).