$a, b, c, d$ are integers with $ad \ne bc$. Show that $1/((ax+b)(cx+d))$ can be written in the form $ r/(ax+b) + s/(cx+d)$. Find the sum $1/1\cdot  4 + 1/4\cdot 7 + 1/7\cdot 10 + ... + 1/2998 \cdot 3001$.
1. **Expressing the given fraction in partial fractions:**

   We start with the equation:
   \[
   \frac{1}{(ax+b)(cx+d)} = \frac{r}{ax+b} + \frac{s}{cx+d}
   \]
   To find \(r\) and \(s\), we multiply both sides by \((ax+b)(cx+d)\):
   \[
   1 = r(cx+d) + s(ax+b)
   \]
   Expanding the right-hand side, we get:
   \[
   1 = rcx + rd + sax + sb
   \]
   Grouping the terms involving \(x\) and the constant terms, we have:
   \[
   1 = (rc + sa)x + (rd + sb)
   \]
   For this equation to hold for all \(x\), the coefficients of \(x\) and the constant terms must be equal. Therefore, we have the system of equations:
   \[
   rc + sa = 0
   \]
   \[
   rd + sb = 1
   \]

2. **Solving the system of equations:**

   From the first equation, we can express \(r\) in terms of \(s\):
   \[
   r = -\frac{sa}{c}
   \]
   Substituting this into the second equation:
   \[
   -\frac{sa}{c}d + sb = 1
   \]
   Simplifying, we get:
   \[
   -\frac{sad}{c} + sb = 1
   \]
   Factoring out \(s\):
   \[
   s\left(b - \frac{ad}{c}\right) = 1
   \]
   Solving for \(s\):
   \[
   s = \frac{1}{b - \frac{ad}{c}} = \frac{c}{bc - ad}
   \]
   Substituting \(s\) back into the expression for \(r\):
   \[
   r = -\frac{sa}{c} = -\frac{a}{bc - ad}
   \]

3. **Verifying the partial fractions:**

   We have:
   \[
   \frac{1}{(ax+b)(cx+d)} = \frac{r}{ax+b} + \frac{s}{cx+d}
   \]
   where:
   \[
   r = -\frac{a}{bc - ad}, \quad s = \frac{c}{bc - ad}
   \]
   Substituting these values back into the original equation confirms the partial fraction decomposition.

4. **Finding the sum:**

   We need to find the sum:
   \[
   \sum_{n=1}^{999} \frac{1}{(3n-2)(3n+1)}
   \]
   Using the partial fraction decomposition:
   \[
   \frac{1}{(3n-2)(3n+1)} = \frac{A}{3n-2} + \frac{B}{3n+1}
   \]
   Solving for \(A\) and \(B\) as before, we get:
   \[
   A = \frac{1}{3}, \quad B = -\frac{1}{3}
   \]
   Therefore:
   \[
   \frac{1}{(3n-2)(3n+1)} = \frac{1}{3} \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right)
   \]
   The sum becomes:
   \[
   \sum_{n=1}^{999} \frac{1}{3} \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right)
   \]
   This is a telescoping series, where most terms cancel out, leaving:
   \[
   \frac{1}{3} \left( 1 - \frac{1}{3001} \right) = \frac{1}{3} \left( \frac{3000}{3001} \right) = \frac{1000}{3001}
   \]

The final answer is \(\boxed{\frac{1000}{3001}}\).