Prove the sum of any $n$ consecutive numbers is divisible by $n$ (when $n$ is odd). 
Let $n \in \mathbb N$ be odd. Prove that the sum of any $n$ consecutive numbers is divisible by $n$.

I started out with $s = x + (x + 1) + (x + 2) + … + (x + n) = kx + n.$ What I am interested in is if that's a right way to sum $x + (x + 1) + (x + 2) + … + (x + n)$ meaning does it equal $kx + n?$

$(x+1)+(x+2)+\cdots+(x+n)=nx+(1+2+\cdots+n)$   
$=nx+\frac{n(n+1)}{2}=n\left(x+\frac{n+1}{2}\right)$. Since $n$ is odd, $x+\frac{n+1}{2}$ is an integer.
