Given
$$\sum\limits_{i=1}^n \left\lfloor \frac{i}{2} \right\rfloor \!\ = \begin{cases} \dfrac{n^2}{4}, & \mbox{if } n\mbox{ is even} \\[1ex] \dfrac{n^2-1}{4}, & \mbox{if } n\mbox{ is odd} \end{cases}$$
for every natural number $n$.
If I put $n=0$ I get $0=0$ but if I put $n=1$ I get $\dfrac{1}{2}=0$. Why?
When $n = 1$, the left-hand-side becomes $\left\lfloor \dfrac12 \right\rfloor = 0$, which is consistent with the right-hand-side. The error occurs because of the omission of the floor sign.