(a) Let $a, b, c, d$ be integers such that $ad\ne bc$. Show that is always possible to write the fraction $\frac{1}{(ax+b)(cx+d)}$in the form $\frac{r}{ax+b}+\frac{s}{cx+d}$

(b) Find the sum $$\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+...+\frac{1}{1995 \cdot 1996}$$
### Part (a)
We need to show that it is always possible to write the fraction $\frac{1}{(ax+b)(cx+d)}$ in the form $\frac{r}{ax+b}+\frac{s}{cx+d}$ for integers $a, b, c, d$ such that $ad \ne bc$.

1. **Assume the form:**
   \[
   \frac{1}{(ax+b)(cx+d)} = \frac{r}{ax+b} + \frac{s}{cx+d}
   \]

2. **Combine the right-hand side using a common denominator:**
   \[
   \frac{r}{ax+b} + \frac{s}{cx+d} = \frac{r(cx+d) + s(ax+b)}{(ax+b)(cx+d)}
   \]

3. **Equate the numerators:**
   \[
   \frac{r(cx+d) + s(ax+b)}{(ax+b)(cx+d)} = \frac{1}{(ax+b)(cx+d)}
   \]
   This implies:
   \[
   r(cx+d) + s(ax+b) = 1
   \]

4. **Expand and group terms:**
   \[
   rcx + rd + sax + sb = 1
   \]
   \[
   (rc + sa)x + (rd + sb) = 1
   \]

5. **Set up the system of equations:**
   For the equation to hold for all $x$, the coefficients of $x$ and the constant terms must match:
   \[
   rc + sa = 0
   \]
   \[
   rd + sb = 1
   \]

6. **Solve the system of linear equations:**
   \[
   \begin{cases}
   rc + sa = 0 \\
   rd + sb = 1
   \end{cases}
   \]

7. **Express $r$ and $s$ in terms of $a, b, c, d$:**
   From the first equation, solve for $r$:
   \[
   r = -\frac{sa}{c}
   \]
   Substitute $r$ into the second equation:
   \[
   -\frac{sa}{c}d + sb = 1
   \]
   \[
   s\left(b - \frac{ad}{c}\right) = 1
   \]
   \[
   s = \frac{c}{ad - bc}
   \]
   Substitute $s$ back into the expression for $r$:
   \[
   r = -\frac{a \cdot \frac{c}{ad - bc}}{c} = \frac{a}{bc - ad}
   \]

8. **Verify the solution:**
   \[
   r = \frac{a}{ad - bc}, \quad s = \frac{c}{ad - bc}
   \]
   These values satisfy the original system of equations, confirming that:
   \[
   \frac{1}{(ax+b)(cx+d)} = \frac{\frac{a}{ad - bc}}{ax+b} + \frac{\frac{c}{ad - bc}}{cx+d}
   \]

### Part (b)
We need to find the sum:
\[
\sum_{n=1}^{665} \frac{1}{(3n-2)(3n+1)}
\]

1. **Express the general term using partial fractions:**
   \[
   \frac{1}{(3n-2)(3n+1)} = \frac{A}{3n-2} + \frac{B}{3n+1}
   \]

2. **Solve for $A$ and $B$:**
   \[
   \frac{1}{(3n-2)(3n+1)} = \frac{A(3n+1) + B(3n-2)}{(3n-2)(3n+1)}
   \]
   \[
   1 = A(3n+1) + B(3n-2)
   \]
   \[
   1 = (3A + 3B)n + (A - 2B)
   \]
   Equate coefficients:
   \[
   3A + 3B = 0 \quad \Rightarrow \quad A = -B
   \]
   \[
   A - 2B = 1
   \]
   Substitute $A = -B$:
   \[
   -B - 2B = 1 \quad \Rightarrow \quad -3B = 1 \quad \Rightarrow \quad B = -\frac{1}{3}, \quad A = \frac{1}{3}
   \]

3. **Rewrite the general term:**
   \[
   \frac{1}{(3n-2)(3n+1)} = \frac{1/3}{3n-2} - \frac{1/3}{3n+1}
   \]

4. **Sum the series:**
   \[
   \sum_{n=1}^{665} \left( \frac{1/3}{3n-2} - \frac{1/3}{3n+1} \right)
   \]
   This is a telescoping series, where most terms cancel out:
   \[
   \frac{1}{3} \left( \frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \cdots + \frac{1}{1993} - \frac{1}{1996} \right)
   \]
   The remaining terms are:
   \[
   \frac{1}{3} \left( 1 - \frac{1}{1996} \right) = \frac{1}{3} \left( \frac{1995}{1996} \right) = \frac{1995}{3 \cdot 1996}
   \]

The final answer is $\boxed{\frac{1995}{3 \cdot 1996}}$