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.
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 $$\left\lfloor \frac{m^2}{4}\right\rfloor=\left\lfloor t^2+t+\frac{1}{4}\right\rfloor=t(t+1).$$