<s>[INST] Prove that the sum $1^k+2^k+\cdots+n^k$ where $n$ is an arbitrary integer and $k$ is odd, is divisible by $1+2+\cdots+n$. 
Prove that the sum $$1^k+2^k+\cdots+n^k$$
  where $n$ is an arbitrary integer and $k$ is odd, is divisible by $1+2+\cdots+n$.

Question
In the solution to this problem it splits it up into two cases: ($1$) $n$ is an even integer ($2$) and $n$ is an odd integer. In the case where $n$ is an odd integer it says the following: $$1^k+n^k,2^k+(n-1)^k,3^k+(n-2)^k,\ldots, \left (\dfrac{n-1}{2} \right )^k + \left(\dfrac{n+3}{2} \right )^k \left (\dfrac{n+1}{2} \right )^k$$ are all divisible by $\dfrac{n+1}{2}$.
I get how the beginning terms are all divisible by $\dfrac{n+1}{2}$, but did they make a typo when they said $\left(\dfrac{n+3}{2} \right )^k \left (\dfrac{n+1}{2} \right )^k$? If not, then how is $\left (\dfrac{n-1}{2} \right )^k + \left(\dfrac{n+3}{2} \right )^k \left (\dfrac{n+1}{2} \right )^k$ divisible by $\dfrac{n+1}{2}$?
 [/INST] Using Proof of $a^n+b^n$ divisible by a+b when n is odd,
$$r^k+(n-r)^k$$ is divisible by $r+n-r=n$  as $k$ is odd
$$\implies\sum_{r=1}^n(r^k+(n-r)^k)$$ will be divisible by $n$
Similarly, 
$$\sum_{r=1}^n(r^k+(n+1-r)^k)$$ will be divisible by $r+n+1-r=n+1$
$$\implies\sum_{r=1}^n(r^k+(n+1-r)^k)=2\sum_{r=1}^n r^k$$ will be divisible by lcm$(n+1,n)$
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