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 ( 1k + 2k + 3k + ... + nk )
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 ( 1k + 2k ) 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 )
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.
1+2 divides 13+23
1+2+3 divides 13+23+33
solve for a. That’s the pattern:
(1+2+3) divides 15+25+35
solve for a.