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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 $\dfrac{n+1}{2}$?

Jacob Willis
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1 Answers1

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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$(n+1,n)$

Community
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lab bhattacharjee
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  • I was also thinking of a similar approach....when I first saw this problem the solution of $a^n+b^n$ is divisible by $a+b$ did click my mind once but I was having difficulty in proving it.You answer is beautiful... – Soham Dec 27 '15 at 13:46