Answer
$$3126250$$
Work Step by Step
$$\eqalign{
& \sum\limits_{j = 1}^{2500} j \cr
& {\text{Use the Summation rule }}\sum\limits_{i = 1}^n i = \frac{{n\left( {n + 1} \right)}}{2}.{\text{ Then,}} \cr
& \sum\limits_{j = 1}^{2500} j = \frac{{2500\left( {2500 + 1} \right)}}{2} \cr
& {\text{Simplify}} \cr
& \sum\limits_{j = 1}^{2500} j = 1250\left( {2501} \right) \cr
& \sum\limits_{j = 1}^{2500} j = 3126250 \cr} $$
