Let S= 1^k +2^k +· · ·+n^k.
Show that S is divisible by 1+2+· · +n for any integer n and odd k.
(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.
re: Proof for n even....for n odd, figure out - Got it!
