Proving that for any odd integer:$\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{(n-1)(n+1)}{4}$ I'm trying to figure out how to prove that for any odd integer, the floor of:
$$\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{(n-1)(n+1)}{4}$$  
Any help is appreciated to construct this proof!  
Thanks guys.

Take $n=2k+1$ then,
$\lfloor(n^2/4)\rfloor=\lfloor k^2+k+1/4\rfloor=k^2+k$
$\frac{(n-1)(n+1)}{4}=(n^2-1)/4=k^2+k=\lfloor(n^2/4)\rfloor$
