Prove that the sum $$1^k+2^k+3^k+...................+n^k.$$
Where 'n' is an arbitrary integer & 'k' is odd, is divisible by "1+2+3+4+5+............+n".
[A question based on 'Divisibility of Integers'.]
An Olympiad question.

Question

maheshmeghwal9 · Answer

@A.Avinash_Goutham @dpaInc @Callisto @satellite73 @Limitless @lgbasallote @myininaya  @kropot72  Plz help:)

maheshmeghwal9 · Answer

@apoorvk

apoorvk · Answer

I guess we need to use the Principle of Mathematical Induction here.

maheshmeghwal9 · Answer

i think too but how?

kinggeorge · Answer

This may or may not work, but all you need to show is that $$2(1+2^k+3^k+...+n^k)$$is divisible by both $n$ and $n+1$.

maheshmeghwal9 · Answer

I have a solution.
but i didn't gt that:(
I m attaching that wait for 2 min.

maheshmeghwal9 · Answer

attachment

kinggeorge · Answer

Interesting. I'll see if I can get somewhere using recursion and some other ideas I have.

anonymous · Answer

@KingGeorge  has it

maheshmeghwal9 · Answer

hmm....
but i m nt getting

maheshmeghwal9 · Answer

I will wait for @KingGeorge

anonymous · Answer

try this http://answers.yahoo.com/question/index?qid=20100227063304AAMpjKX

maheshmeghwal9 · Answer

So using Arithmetic Progression & Mathematical Induction can we do it?