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 value of `(b-2d)/(8)` is

Options

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

Answer

To solve the problem step by step, we need to analyze the sequence of odd integers given and find the values of \( b \), \( c \), and \( d \) such that the equation \( a_n = b[\sqrt{n+c}] + d \) holds for the sequence.

### Step 1: Understand the sequence
The sequence provided is:
\[ a_1 = 1, a_2 = 3, a_3 = 3, a_4 = 5, a_5 = 5, a_6 = 5, a_7 = 5, a_8 = 5, \ldots \]
This means that each positive odd integer \( k \) appears \( k \) times.

### Step 2: Identify the pattern
From the sequence:
- 1 appears 1 time,
- 3 appears 3 times,
- 5 appears 5 times,
- 7 appears 7 times,
- and so on.

### Step 3: Determine the values of \( b \), \( c \), and \( d \)
We know that:
\[ a_n = b[\sqrt{n+c}] + d \]
We can observe that:
- For \( n = 1 \), \( a_1 = 1 \)
- For \( n = 2 \), \( a_2 = 3 \)
- For \( n = 3 \), \( a_3 = 3 \)
- For \( n = 4 \), \( a_4 = 5 \)

Let’s analyze the values:
1. For \( n = 1 \):
\[ a_1 = 1 = b[\sqrt{1+c}] + d \]
2. For \( n = 2 \):
\[ a_2 = 3 = b[\sqrt{2+c}] + d \]
3. For \( n = 3 \):
\[ a_3 = 3 = b[\sqrt{3+c}] + d \]
4. For \( n = 4 \):
\[ a_4 = 5 = b[\sqrt{4+c}] + d \]

### Step 4: Solve for \( b \) and \( d \)
From the first equation:
\[ 1 = b[\sqrt{1+c}] + d \]
From the second equation:
\[ 3 = b[\sqrt{2+c}] + d \]

Subtracting these two equations gives:
\[ 3 - 1 = b[\sqrt{2+c}] - b[\sqrt{1+c}] \]
\[ 2 = b(\sqrt{2+c} - \sqrt{1+c}) \]

### Step 5: Find \( c \)
Now, we can set \( b = 2 \) (as it fits the pattern of odd integers) and solve for \( c \):
\[ 2 = 2(\sqrt{2+c} - \sqrt{1+c}) \]
This simplifies to:
\[ 1 = \sqrt{2+c} - \sqrt{1+c} \]
Squaring both sides:
\[ 1 = (2+c) + (1+c) - 2\sqrt{(2+c)(1+c)} \]
This leads to:
\[ 2\sqrt{(2+c)(1+c)} = 3 + c \]
Squaring again and solving will yield \( c = -1 \).

### Step 6: Find \( d \)
Using \( b = 2 \) and \( c = -1 \) in the first equation:
\[ 1 = 2[\sqrt{1-1}] + d \]
This gives:
\[ 1 = d \implies d = 1 \]

### Step 7: Calculate \( \frac{b - 2d}{8} \)
Now we can calculate:
\[ \frac{b - 2d}{8} = \frac{2 - 2 \cdot 1}{8} = \frac{2 - 2}{8} = \frac{0}{8} = 0 \]

### Final Answer
Thus, the possible value of \( \frac{b - 2d}{8} \) is \( 0 \).