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I want to compute the following sum $$S = \sum_{k=0}^{m} \left\lfloor \frac{k}{2}\right\rfloor.$$ Here is what I tried: $$ S = \sum_{k\geq 0, 2|k}^{m} \left\lfloor \frac{k}{2}\right\rfloor + \sum_{k\geq 0, 2\not |k}^{m} \left\lfloor \frac{k}{2}\right\rfloor.$$ If $m= 2t$ then $$S =\sum_{k\geq 0, 2|k}^{m} \left\lfloor \frac{k}{2}\right\rfloor + \sum_{k\geq 0, 2\not |k}^{m} \left\lfloor \frac{k}{2}\right\rfloor = \frac{t(t+1)}{2} + \frac{(t-1)t}{2} = t^2.$$ If $m= 2t+1$ then $$S = \sum_{k\geq 0, 2|k}^{m} \left\lfloor \frac{k}{2}\right\rfloor + \sum_{k\geq 0, 2\not |k}^{m} \left\lfloor \frac{k}{2}\right\rfloor = \frac{t(t+1)}{2} + \frac{t(t+1)}{2}= t(t+1).$$

But I am not sure if this is correct. Perhaps someone could give an indication.

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    $\begingroup$ You can write $t$ in terms of $m$ in each case. $\endgroup$ – model_checker Jan 11 at 8:35
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    $\begingroup$ @Hello_World Actuallly, you can. $\endgroup$ – 5xum Jan 11 at 8:38
  • $\begingroup$ Are you asking me to write them in terms of $m$? I can do that if it helps. $\endgroup$ – model_checker Jan 11 at 8:39
  • $\begingroup$ @KemonoChen, please write that as an answer so that we can downvote it. $\endgroup$ – Carsten S Jan 11 at 14:42
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Yes, you are correct. You may also write the result as a more compact formula: $$\sum_{k=0}^{m} \left\lfloor \frac{k}{2}\right\rfloor= \begin{cases} t^2&\text {if $m=2t$}\\ t(t+1)&\text {if $m=2t+1$}\\ \end{cases}=\left\lfloor \frac{m^2}{4}\right\rfloor.$$ Indeed, if $m=2t$ then $$\left\lfloor \frac{m^2}{4}\right\rfloor=\left\lfloor t^2\right\rfloor=t^2$$ and if $m=2t+1$ then $$\left\lfloor \frac{m^2}{4}\right\rfloor=\left\lfloor t^2+t+\frac{1}{4}\right\rfloor=t(t+1).$$

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