Series

$\begin{array}{ccl} \displaystyle\sum_{k=1}^{500} (2k-1)^3 & = & \displaystyle\sum_{m=1}^{1000} m^3 - \displaystyle\sum_{n=1}^{500} (2n)^3 \\ & = & \displaystyle\sum_{m=1}^{1000} m^3 - 8 \displaystyle\sum_{n=1}^{500} n^3 \\ & = & \left(\dfrac{1000*1001}{2} \right)^2 - 8 \left(\dfrac{500*501}{2} \right)^2 \\ & = & 500^2 *1001^2 - 8 * 250^2 *501^2 \\ & = & 124999750000 \end{array}$