$$\sum\limits_{i=1}^n \left\lfloor \frac{i}{2} \right\rfloor \!\ = \begin{cases} \dfrac{n^2}{4}, & \mbox{if } n\mbox{ is even} \\[1ex] \dfrac{n^2-1}{4}, & \mbox{if } n\mbox{ is odd} \end{cases}$$

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?

  • $\begingroup$ If $n=1$, you also have $0=0$. $\endgroup$ – Bill O'Haran May 11 '18 at 12:54
  • 2
    $\begingroup$ As was mentioned on your other post, $\big\lfloor\dfrac{1}{2}\big\rfloor=0$, so you get the expected result. $\endgroup$ – B. Mehta May 11 '18 at 12:54
  • $\begingroup$ If you want to allow $n=0$, it would make sense to start the sum at $i=0$ rather than $i=1$. $\endgroup$ – Barry Cipra May 11 '18 at 15:52

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.

  • $\begingroup$ How can I prove that $\frac{n^2}{4}=\frac{(n+2)^2}{4}$? $\endgroup$ – user557276 May 11 '18 at 13:03
  • $\begingroup$ @user557276 No, you can't. Consider the case when $n=0$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 11 '18 at 13:04
  • $\begingroup$ The exercise tells me to prove that $P(n)→P(n+2)$ $\endgroup$ – user557276 May 11 '18 at 13:06
  • $\begingroup$ @user557276 What is $P(n)$? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 11 '18 at 13:09
  • $\begingroup$ It's the original expression. I have to prove that $P(n)→P(n+2)$ first when $n$ is even and then when $n$ is odd $\endgroup$ – user557276 May 11 '18 at 13:11

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