Question

A quickie problem for someone to answer. Medals are up on the plate.

Answer 1

Let's say that there is a series of numbers S such that \[S = 1^k + 2^k + 3^k.............n^k\]. Prove the following:

S is divisible by \[(1 + 2 + 3 + 4 + ....................n)\]

Answer 2

Note: k is a natural number.

Answer 3

@hoblos @myininaya @mathslover @Shane_B @sasogeek @Australopithecus @Libniz @DLS @perl

Answer 4

Let k =2 n=3
\[1^2+2^2+3^2 = 14\]Not divisible by 1+2+3=6 So maybe k > 2?

Answer 5

Ah yes, sorry. k should be an odd number.

Answer 6

Correction: "such k that it is an odd number."

Answer 7

\[\large S=\sum_{i=1}^ni^{2k-1}\]\[\large\text{Let }R=\sum_{i=1}^n{i}=\frac{n(n+1)}{2}\]Show R divides S for k natural.\[\large\text{Let }Q=(\sum_{i=1}^n{i})^{2k-1}=\Bigg(\frac{n(n+1)}{2}\Bigg) ^{2k-1}=R^{2k-1}\] So R divides Q. If R also divides Q-S, then R divides S. So what is Q-S?...