Oh.. I thought this was your homework. The good thing that you have a passion to help whoever that person, but I think I'm late to answer this one. Sorry dude.
n and k are natural numbers, k is odd
Prove that ( 1 + 2 + 3 + ... + n ) is a factor of ( 1^{k} + 2^{k} + 3^{k} + ... + n^{k} )
I start the proof by using mathematical induction ( but don't know how to do it ), then I try to tackle from a special case :
[ e.g. The case ( 1 + 2 ) is a factor of ( 1^{k} + 2^{k} ) where k is natural and odd --- I can proof it but do not know how to extend it ]
Can anyone give me a suggestion ( Better be simple because I am not good at math. )
** This is not a homework. I saw this question on another web site ( not a math. forum ) 2 days ago.
I thought it was an easy question so I have spent the past 2 days to solve it. ( It seems I was overly confident )
JeanGunter, you have repeatedly posted a response saying you cannot help! Why? There are, I am sure, millions of people who cannot help with a particular problem. If they all posted to say so, this board would be overwhelmed!
AnEducatedMonkey, you can't prove it. It is NOT true. In particular, if n= 2 and i= 2, this asserts that 1+ 2= 3 must divide and that obviously is not true.
he does specify k (what I think you mean by i) is odd. It does seem to work for k odd. I took a quick look. Unless I missed something it's a non-trivial problem. Specifying the sum of 1 to n is simple enough, but for general odd k the second sum involves the generalized Harmonic number (over my head). Info can be found by googling "power sum".
1+2 divides 1^{3}+2^{3}
(1+2)(2^{2}+a)=2^{2}+a+2^{3}+2a=1+2^{3}
a=-1
ck:
(1+2)(2^{2}-1)=1^{3}+2^{3}
1+2+3 divides 1^{3}+2^{3}+3^{3}
(1+2+3)(2^{2}+3^{2}+a)= 1+2^{3}+3^{3}
solve for a. That’s the pattern:
(1+2+3) divides 1^{5}+2^{5}+3^{5}
(1+2+3)(2^{4}+3^{4}+a)= 1+2^{5}+3^{5}
solve for a.
etc, etc