Let S= 1^k +2^k +· · ·+n^k.
Show that S is divisible by 1+2+· · +n for any integer n and odd k.
(a^k + b^k) is divisible by (a + b), k odd.
1 + 2 + 3 + ... + n = n * (n + 1)/2

First part: n is an even integer.
There´s an even number of terms in S. Group them this way:
[1^k + n^k] + [2^k + (n1)^k] + ...
Each sum enclosed by brackets is divisible by (n + 1), so the whole S is divisible by (n + 1).
Grouping S this way:
[1^k + (n1)^k] + [2^k + (n2)^k] + ... + [n/2]^k + [n^k]
Each sum enclosed by brackets is divisible by n, so by n/2.
Thus, for n even, S is divisible by (n+1) and also by (n/2), thus by n*(n+1)/2.

Second part: n is an odd integer.
There´s an odd number of terms in S.
Figure out how to group them...

Posted by pcbouhid
on 20100726 15:11:01 