Number of Quadratic equation with different condition If the number of quadratic polynomials $ax^2+2bx+c$  which satisfy the following conditions:
(i) a, b, c are distinct
(ii) a, b, c $\in$ ${1, 2, 3,. 2001, 2002}$
(iii) x + 1 divides $ax^2+2bx+c$
is equal to $1000 \lambda$, then find the value of $\lambda$.
My solution is as follow
x + 1 divides hence $f(x)=ax^2+2bx+c$, hence $f(-1)=0$, therefore $2b=a+c$
As a.b.c are distinct therefore $a\ne c$
Maximum value of  b is 1000 and minimum value of b is 1 and all the value increases by 1
Minimum value of a+c=2 and Maximum value of a+c=2000
It is like distribution of 2b identical balls into 2 distinct boxes so that each box has at least 1 balls ${}^{2b - 1}{C_{2 - 1}} \Rightarrow {}^{2b - 1}{C_1}$
b=1; ${}^{2 - 1}{C_{2 - 1}} = 1$
b=2; ${}^{4 - 1}{C_{2 - 1}} = 3$
b=1000, ${}^{2000 - 1}{C_{2 - 1}} = 1999$
The total number of ways=$\frac{n}{2}\left( {2a' + \left( {n - 1} \right)d'} \right) = \frac{{1000}}{2}\left( {2 + \left( {1000 - 1} \right)2} \right) = {1000^2}$
Now as a,b and c are distinct removing the following cases {1,1},{2,2},....{1000,1000}
Hence total number of cases are ${\left( {1000} \right)^2} - 1000 = 1000 \times 999, \lambda  = 999$ but answer is $2002$, i cannot figure out my mistake

Let us understand the condition $a+c=2b$, where $a,b,c$ are natural numbers. Here $2b$ is even, which can be obtained only by adding two numbers of same parity. Among first $2002$ natural numbers, there are $1001$ odd and $1001$ even numbers. Hence number of ordered pairs $(a,c)$ should be given by
$$2\times \left( \binom{1001}{2}+\binom{1001}{2} \right) = 1000 \times 2002$$
