Find
\(\displaystyle \sum_{r = 1}^{100} \left\lfloor \frac{1}{2} + \frac{r}{100} \right\rfloor\)
Note: \(\lfloor x \rfloor\) stands for greatest integer (floor) function
sumfor(r, 1, 100, floor(r/100 + 1/2)) = 51
