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 Oddly divisible (Posted on 2010-07-17)
Let S= 1^k +2^k +· · ·+n^k.
Show that S is divisible by 1+2+· · +n for any integer n and odd k.

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 Proof for n even....for n odd, figure out | Comment 4 of 5 |
(a^k + b^k) is divisible by (a + b), k odd.

1 + 2 + 3 + ... + n = n * (n + 1)/2

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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 + (n-1)^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 + (n-1)^k] + [2^k + (n-2)^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.

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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 2010-07-26 15:11:01

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