Not an AP or GP

$\begin{aligned} S & = \frac {1^2}{1 \cdot 3} + \frac {2^2}{3 \cdot 5} + \frac {3^2}{5 \cdot 7} + ... + \frac {500^2}{999 \cdot 1001} \\ & = \sum_{n=1}^{500} \frac {n^2}{(2n-1)(2n+1)} \\ & = \sum_{n=1}^{500} \frac {4n^2-1+1}{4(2n-1)(2n+1)} \\ & = \frac 14 \sum_{n=1}^{500} \frac {(2n-1)(2n+1)+1}{(2n-1)(2n+1)} \\ & = \frac 14 \sum_{n=1}^{500} \left( 1 + \frac 1{(2n-1)(2n+1)} \right) \\ & = \frac 14 \sum_{n=1}^{500} \left( 1 + \frac 12 \left[ \frac 1{2n-1} - \frac 1{2n+1} \right] \right) \\ & = \frac 14 \left( 500 + \frac 12 \left[ \frac 11 - \frac 1{1001} \right] \right) \\ & = \boxed{\dfrac {125250}{1001}} \end{aligned}$

S= $\frac{1}{2}$ ( $\frac{1}{1}$ - $\frac{1}{3}$ )+ $\frac{4}{2}$ ( $\frac{1}{3}$ - $\frac{1}{5}$ )............................... $\frac{250000}{2}$ ( $\frac{1}{999}$ - $\frac{1}{1001}$ )

S=( $\frac{1}{2}$ {- $\frac{1}{6}$ + $\frac{4}{6}$ }{- $\frac{4}{10}$ + $\frac{9}{10}$ }...................................)

S=( $\frac{1}{2}$ ........................500 terms - $\frac{250000}{2}$ ( $\frac{1}{1001}$ )

S= $\frac{500}{2}$ - $\frac{250000}{2002}$

S= $\frac{500500-250000}{2002}$

S= $\frac{250500}{2002}$

S= $\frac{125250}{1001}$