How can I prove that $\sum\limits_{m=k}^{k+2n}\!\!\!m$ is divisible by $2n + 1$? Show that $\sum\limits_{m=k}^{k+2n}\!\!\!m$ is divisible by $2n + 1$, where $n > 0$ and $k > 0$.
I don't know how to go about this question, any help will be greatly appreciated.
 A: A non-computational solution:  Considered modulo $(2n+1)$, the numbers $k, k+1, \ldots, k+2n$ form a complete residue system.  In computing the sum of these, each number pairs with its negative.  Because $2n+1$ is odd, there is no number that is its own negative, hence the sum is zero, modulo $2n+1$, and hence $2n+1$ divides the sum.
A: The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$
$\displaystyle \sum_{m=k}^{k+2n}m=\frac{(k+2n)(k+2n+1)-k(k-1)}2=\frac{4nk+4n^2+2n+2k}2=(2n+1)(n+k)$
A: HINT: $$\begin{align}
\sum_{m=k}^{k+2n}m &= \sum_{m=k}^{k+2n}(m-k+k) \\
&= \sum_{m=k}^{k+2n}(m-k)+\sum_{m=k}^{k+2n}k
\end{align}$$
Now, $$\begin{align}
\sum_{m=k}^{k+2n}(m-k) &= \sum_{i=0}^{2n}i=\frac{2n(2n+1)}{2}\\
&=n(2n+1)\qquad{\textstyle \left(\sum_{i=1}^n=\frac{n(n+1)}{2}\right)}
\end{align}$$
and $$\sum_{m=k}^{k+2n}k=k(2n+1).$$
A: Using the formula of the summation of Arithmetic Series 
$\sum_{0\le r\le n-1}(a+r\cdot d)=\frac n2\{ {2a+(n-1)d}\}$ where $a$ is the first term, $n$ is the number of terms, and $d$ is the common difference
$$\sum\limits_{m=k}^{k+2n}m=\frac{(2n+1)}2\{2\cdot k+(2n+1-1)\cdot1\}=(2n+1)(n+k)$$
A: $\displaystyle \sum_{m=k}^{k+2n}m=k(2n+1) + \sum_{m=0}^{2n}m$  because m ≥k for every one of the 2n+1 terms (in fact, we could say that m = k + a for an integer a between 0 and 2n)! Also, that last summation is easily evaluated as m(m+1)/2. Plug in the values of m and Q.E.D.
