USSR OLYMPIAD PROBLEM
-
MATHPRITOM
- Posts:190
- Joined:Sat Apr 23, 2011 8:55 am
- Location:Khulna
Prove that,$1^k+2^k+3^K+...+n^k$,where n is an arbitrary integer & k is odd,is divisible by 1+2+3+...+n.
Re: USSR OLYMPIAD PROBLEM
Note that $n|1^k+(n-1)^k$
And $n+1|1^k+n^k$
And $n+1|1^k+n^k$
One one thing is neutral in the universe, that is $0$.