Suspicion: It's divisible by 3 because it's 3 consecutive integers. How about 3, 4, 5: adds up to 12. 4, 5, 6: adds up to 15.

However, it only seems to work for an odd count of consecutive integers, e.g. 2 + 3 + 4 + 5 = 14, not divisible by 4.

If we regard the **median** as the symmetry axis:

2 + **3** + 4

n-1 + **n** + n+1; -1 +1 cancel each other out, we're left with n + **n** + n = 3n.

3 + 4 + **5** + 6 + 7

n-2 + n-1 + **n** + n+1 + n+2; -2 +2, -1 +1, cancel each other out, we're left with n + n + **n** + n + n = 5n.

Conjecture: **The sum of k consecutive integers with median n is kn if k is odd.**
Proof below.

4 5 6 => k = 3, n = 5, kn = 15.

k | 3 | 3 | 3 | 3 | 3 | ||||||||||

n | 5 | 5 | 5 | ||||||||||||

kn | 4 | 5 | 6 |

k | 3 | 3 | 3 | 3 | 3 | 3 | 3 | ||||||||||||||

n | 7 | 7 | 7 | ||||||||||||||||||

kn | 6 | 7 | 8 |

2 3 4 5 6 => k = 5, n = 4, kn = 20.

k | 5 | 5 | 5 | 5 | ||||||||||||||||

n | 4 | 4 | 4 | 4 | 4 | |||||||||||||||

kn | 2 | 3 | 4 | 5 | 6 |

The sum of the first n integers is m(m+1)/2.
So the sum of k consecutive integers is

m(m+1)/2 - (m-k)(m-k+1)/2, simplified:

((2m+1)k - k^{2})/2. According to the conjecture this is supposed to equal kn if k is odd.

((2m+1)k - k^{2})/2 = kn.

(2m+1 - k)/2 = n.

n must be integer, so (2m+1 - k)/2 must be integer, too, hence (2m+1 - k) must be even.

2m+1 is guaranteed to be odd, hence k must be odd in order for the result to be even.

q.e.d.

© Bernhard Wagner, May 28th, 2012.