Mosco Olympiad (Part I)

We note that both $m^2-m+1$ and $m^2+m-1$ are odd numbers, with $m^2-m+1 < m^2+m-1$ for $m \in \mathbb N$ . The sequence of odd numbers from $m^2-m+1$ to $m^2+m-1$ is an arithmetic progression with the first term $a=m^2-m+1$ , common difference $d=2$ and last term $l = m^2+m-1$ .

Therefore, the number of terms is

$\begin{aligned} n & = \frac {l-a}2+1 \\ & = \frac {m^2+m-1-(m^2-m+1)}2+1 \\ & = \frac {2m-2}2+1 \\ & = m \end{aligned}$

The sum of all odd numbers is

$\begin{aligned} S & = \frac {n(a+l)}2 \\ & = \frac {m(m^2+m-1+m^2-m+1)}2 \\ & = \frac {2m^3}2 \\ & = \boxed{m^3} \end{aligned}$