Prove the sum of any $n$ consecutive numbers is divisible by $n$ (when $n$ is odd).

Question

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 $kx + n?$

Answer 1

$(x+1)+(x+2)+\cdots+(x+n)=nx+(1+2+\cdots+n)$

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

Answer 2

Note that you sum up $n+1$ numbers.

Let $n=2m-1$. $$\sum_{i=1}^{2m-1}(x+i-1)=(2m-1)(x-1)+\frac{(2m-1)(2m)}{2}=(2m-1)(x-1+m).$$

Answer 3

$\bmod n\!:\:\:\: 0+1+2+\cdots+(n-1)$

$\equiv 0+(1+(n-1))+(2+(n-2))+\cdots+(\frac{n-1}{2}+\frac{n+1}{2})$

$\equiv 0+(1-1)+(2-2)+\cdots+(\frac{n-1}{2}-\frac{n-1}{2})\equiv 0$

Since $n$ is odd, we have residues with their negatives in pairs.