Sum of numbers between consecutive multiple numbers of $N$ proof

I need to see if I can generalize a proof: whether the sum of all numbers between two consecutive numbers multiples of $N$, being $N$ a natural number such that $N > 2$ is a multiple of $N$.

I started by taking two consecutive multiples of $N$:

$N$, and $Nn + N$

The sum of the numbers between them is:

$\ (Nn + 1) + (Nn + 2) + (Nn + 3) \ldots (Nn + (N - 3)) + (Nn + (N - 2)) + (Nn + (N - 1)) = N^2n + N$

But

$\ N^2n + N = {N(Nn + 1)}$

Edit:

I didn't sum correctly, the actual sum is:

$\ (2Nn+N)(N-1)/2 $

So the sum of all the numbers between the consecutive multiples of $N$, is a multiple of $N$.

I don't know if the process so far is good, and another problem I have is how to include the fact that $N$, has to be a natural number bigger than 2. How should I go about this?

Thanks in advance.

Answer

Your two consecutive multiples of $N$ should be $Nn$ and $Nn+N$. The sum of all the numbers between them is then $(N-1)Nn+\sum_{i=1}^{N-1}i=N(N-1)n+\frac 12(N-1)N$-you lost the factor $\frac 12$ If $2$ does not divide into $N-1$ your proposition will fail. For example, let $N=4, n=3$. The numbers between $12$ and $16$ are $13,14,15$, with sum $42$, which is not divisible by $4$