To multiply or not to multiply

let $\frac{1}{4×9}$ = $\frac{1}{n(n+5)}$

$\frac{1}{n(n+5)}$ = $\frac{A}{n}$ + $\frac{B}{n+5}$

1=A(n+5)+B(n)

Substituting n=0 and n=-5 gives us A= $\frac{1}{5}$ and B=- $\frac{1}{5}$

Substituting A and B back will give us $\frac{1}{5n}$ - $\frac{1}{5n+25}$

Now the expression becomes

$\frac{1}{4×9}$ + $\frac{1}{9×14}$ + $\frac{1}{70}$ - $\frac{1}{95}$ + $\frac{1}{95}$ - $\frac{1}{120}$ + $\frac{1}{120}$ - $\frac{1}{145}$ +...+ $\frac{1}{9970}$ - $\frac{1}{9995}$ + $\frac{1}{9995}$ - $\frac{1}{10020}$

where the center part telescopes and what's left would be

$\frac{1}{20}$ - $\frac{1}{10020}$ = $\boxed{\frac{25}{501}}$

Another solution:

$\dfrac{1}{4 \times 9} + \dfrac{1}{9 \times 14}\\ = \dfrac{14}{4 \times 9 \times 14} + \dfrac{4}{4 \times 9 \times 14}\\ = \dfrac{18}{4 \times 9 \times 14}\\ = \dfrac{2}{4 \times 14}$

Next, we add the third term:

$\dfrac{2}{4 \times 14} + \dfrac{1}{14 \times 19}\\ = \dfrac{38}{4 \times 14 \times 19}+ \dfrac{4}{4 \times 14 \times 19}\\ = \dfrac{42}{4 \times 14 \times 19}\\ = \dfrac{3}{4 \times 19}$

Have you noticed a pattern? If yes, good for you. The sum of the expression

$\dfrac{1}{4 \times 9} + \dfrac{1}{9 \times 14} + \dfrac{1}{14 \times 19} + \ldots + \dfrac{1}{(5n-1)(5n+4)}$

where $n$ is the number of terms in the expression, is equivalent to

$\dfrac{n}{4 \times (5n+4)}$

Now, for this particular sum, the total is given as

$\dfrac{n}{4 \times 2004}$

We need to find the value of $n$ :

$2004 = 5n +4\\ 5n=2000\\ n=400$

Substitute it in:

$\dfrac{400}{4 \times 2004} = \dfrac{100}{2004} = \dfrac{25}{501}$

Therefore,

$\dfrac{1}{4 \times 9} + \dfrac{1}{9 \times 14} + \dfrac{1}{14 \times 19} + \ldots + \dfrac{1}{1999 \times 2004}= \boxed{\dfrac{25}{501}}$