Question

Prove that if n is an odd integer, then$⌈n^2/4⌉=\frac{\left(n^2+3\right)}{4}$

Short answer

$⌈n^2/4⌉=\frac{\left(n^2+3\right)}{4}$

Step-by-step solution

Step: 1

Write n=2k+1 for some integer k.

Then$n^2=4k^2+4k+1$ ,

so $\frac{n^2}{4}=k^2+k+\frac{1}{4}$.

Step: 2

Therefore,$\left[\frac{n^2}{4}\right]=k^2+k+1$

Step: 3

But

$$\begin{gathered} \frac{\left(n^2+3\right)}{4}=\frac{\left(4k^2+4k+1+3\right)}{4} \\ =k^2+k+1 \end{gathered}$$

Therefore$⌈n^2/4⌉=\frac{\left(n^2+3\right)}{4}$