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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

 Jun 9, 2020
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sumfor(r, 1, 100, floor(r/100 + 1/2)) = 51

 Jun 9, 2020

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