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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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 re: Proof for n even....for n odd, figure out - Got it! Comment 5 of 5 |
(In reply to Proof for n even....for n odd, figure out by pcbouhid)

Your proof for the divisibility into n+1 for even n can be modified for odd n.  Note that (n+1)/2 is an integer in this case.

Group as before.  There will be (n-1)/2 pairs and one lone term:
[1^k + n^k] + [2^k + (n-1)^k] + ... + [((n-1)/2)^k + ((n+3)/2)^k] + [((n+1)/2)^k]

Each of the pairs is divisible by n+1 and the lone term is a power of (n+1)/2.  Therefore the entire sum is divisible by (n+1)/2.

Then for divisibility by n:
[1^k + (n-1)^k] + [2^k + (n-2)^k] + ... + [((n-1)/2)^k + ((n+1)/2)^k]^k + [n^k]

Each of these pairs is divisible by n and lone term is a power of n. Therefore the entire sum is divisible by n.

These two divisibilities combined prove the n is odd case.

 Posted by Brian Smith on 2016-06-27 11:55:49

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