Answer
Let $n$ be any odd integer. Then $n=2k+1$ for some integer $k$, so $\lceil\frac{n^{2}}{4}\rceil=\lceil\frac{(2k+1)^{2}}{4}\rceil=\lceil\frac{4k^{2}+4k+1}{4}\rceil$$=\lceil k^{2}+k+\frac{1}{4}\rceil$. Applying the analogue of ceiling functions for Theorem 4.5.1, we get $\lceil k^{2}+k+\frac{1}{4}\rceil=k^{2}+k+\lceil\frac{1}{4}\rceil=k^{2}+k+1$. Substituting $\frac{n-1}{2}$ for $k$, we in turn get $k^{2}+k+1$$=(\frac{n-1}{2})^{2}+\frac{n-1}{2}+1=\frac{(n^{2}-2n+1)+(2n-2)+4}{4}=\frac{n^{2}+3}{4}$. Since $n$ was an arbitrary odd integer, we conclude that the result holds for all odd integers.
Work Step by Step
By "analogue for ceiling functions of Theorem 4.5.1," we mean that for any integer $m$ and any real number $k$, it is the case that $\lceil m+x\rceil=m+\lceil x\rceil$. The proof of this is quite simple and is almost identical to the proof of Theorem 4.5.1 itself.
An alternative would be to make use of the "weaker" result of exercise 7, which states the same for $0\leq x\lt1$.
