The number of quadratic polynomials `ax^(2)+2bx+c` which satisfy the following conditions is k
(i) a, b, c are distinct
(ii) `a, b, c in {1, 2, 3, 4,….2001, 2002}`
(iii) `x+1` divides `ax+2bx+c` Then `(k)/(10^(5))` is equal to

Answer

To solve the problem, we need to find the number of distinct quadratic polynomials of the form \( ax^2 + 2bx + c \) that satisfy the given conditions. Let's break down the solution step by step.

### Step 1: Understand the conditions
We have the following conditions:
1. \( a, b, c \) are distinct.
2. \( a, b, c \) can take values from the set \( \{1, 2, \ldots, 2002\} \).
3. \( x + 1 \) divides \( ax^2 + 2bx + c \).

### Step 2: Use the divisibility condition
The condition that \( x + 1 \) divides \( ax^2 + 2bx + c \) means that substituting \( x = -1 \) into the polynomial should yield zero:
\[ a(-1)^2 + 2b(-1) + c = 0 \implies a - 2b + c = 0 \implies c = 2b - a. \]

### Step 3: Determine the range for \( a \) and \( b \)
From the equation \( c = 2b - a \), we need to ensure that \( c \) is also in the range \( \{1, 2, \ldots, 2002\} \). This gives us the following inequalities:
1. \( 1 \leq c \leq 2002 \)
2. \( 1 \leq 2b - a \leq 2002 \)

From the second inequality:
- \( 2b - a \geq 1 \implies a \leq 2b - 1 \)
- \( 2b - a \leq 2002 \implies a \geq 2b - 2002 \)

### Step 4: Determine the valid range for \( a \)
Combining these inequalities, we have:
\[ 2b - 2002 \leq a \leq 2b - 1. \]
Since \( a \) must also be a distinct integer in the range \( \{1, 2, \ldots, 2002\} \), we must ensure:
- \( 2b - 2002 \geq 1 \implies b \geq 1002 \)
- \( 2b - 1 \leq 2002 \implies b \leq 1001 \)

However, since \( b \) must also be distinct from \( a \) and \( c \), we can only take values of \( b \) from \( 1 \) to \( 1001 \).

### Step 5: Count the valid pairs \( (a, b) \)
For each valid \( b \) from \( 1 \) to \( 1001 \):
- The value of \( a \) can take values from \( \max(1, 2b - 2002) \) to \( \min(2002, 2b - 1) \).
- The number of valid \( a \) values is given by:
\[ \text{Number of valid } a = \min(2002, 2b - 1) - \max(1, 2b - 2002) + 1. \]

### Step 6: Calculate the total number of distinct polynomials
For each \( b \), we compute the number of valid \( a \) values and then multiply by the number of choices for \( c \) (which is determined by \( c = 2b - a \)).

### Step 7: Calculate \( k \)
After calculating the total number of valid pairs \( (a, b) \), we find \( k \) and then compute \( \frac{k}{10^5} \).

### Final Calculation
The final answer will be \( \frac{k}{10^5} \).