# 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 $k > 0$.

• Welcome to MSE. Regarding your question, I have no idea what "k+2n E m=k m" means. Jun 30 '13 at 4:08
• I think it means $\displaystyle \sum_{m=k}^{k+2n} m$. @user2162882: do you know how to compute $\displaystyle \sum_{i=1}^n i$? Jun 30 '13 at 4:13
• @QiaochuYuan Nice guess. Jun 30 '13 at 4:16
• @QiaochuYuan: Great reading of the question. Jun 30 '13 at 4:18
• @QiaochuYuan, no, and thanks for understanding what I meant in the question. Jul 1 '13 at 0:51

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.

• That would seem to generalize to $2jn+1$ consecutive terms for any positive integer $j$. Jun 30 '13 at 5:13
• Sure, and if the middle term is a multiple of $2n+1$, then you can have any odd number of consecutive terms. Jun 30 '13 at 5:17

The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$

$\displaystyle \sum_{m=k}^{k+2n}m=\frac{(k+2n)(k+2n+1)-k(k-1)}2=\frac{4nk+4n^2+2n+2k}2=(2n+1)(n+k)$

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

$$\sum\limits_{m=k}^{k+2n}m=\frac{(2n+1)}2\{2\cdot k+(2n+1-1)\cdot1\}=(2n+1)(n+k)$$

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 $$\sum_{m=k}^{k+2n}k=k(2n+1).$$

$\displaystyle \sum_{m=k}^{k+2n}m=k(2n+1) + \sum_{m=0}^{2n}m$ 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.