powers sums and triangular numbers

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#1 h Apr 17, 2025, 1:00 PM

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prove 1^k+2^k+3^k + \cdots + n^k \text{is divisible by } \frac{n(n+1)}{2} \text{when} k \text{is odd}

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#2 h Apr 17, 2025, 1:02 PM

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how do i use latex here?

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#3 h Apr 17, 2025, 4:37 PM • 1 Y

Y by gaussious

@above Put dollar signs around your text; for example, $1+1=2$ would give $1+1=2$ . Unfortunately, I think you need to be on aops for at least 2 weeks to do that.

This post has been edited 2 times. Last edited by Quidditch, Apr 18, 2025, 5:11 AM

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#4 h Apr 17, 2025, 4:49 PM • 1 Y

Y by gaussious

gaussious wrote:

prove $1^k+2^k+3^k + \cdots + n^k \text{ is divisible by } \frac{n(n+1)}{2} \text{when } k \text{ is odd}$

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#5 h Apr 18, 2025, 8:38 PM

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This may help.

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#6 h Apr 18, 2025, 10:27 PM

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We essentially want to show that $n(n+1)\mid 2(1^k+2^k+\cdots +n^k)$ , as $gcd(n,n+1)$ it suffices to show $n\mid 2(1^k+2^k+\cdots +n^k)$ and $n+1\mid 2(1^k+2^k+\cdots +n^k)$ .
We first show that $n+1$ divides our sum. If $n$ is even, consider the pairings $(1,n), (2,n-1) \cdots (\frac{n}{2},\frac{n+2}{2})$ . As $k$ is odd, we can factorize each of these terms into being divisible by $n$ . Similarly, if $n$ is odd, consider the same pairings, except we end up with $\frac{n+1}{2}$ not in a pair, but because of the factor of $2$ , this term also ends up being divisible by $n+1$ . Thus, $n+1\mid 2(1^k+2^k+\cdots +n^k)$ .
Now, assume $n$ is odd. Consider the pairs $(1,n-1),(2,n-2)\cdots (\frac{n-1}{2},\frac{n+1}{2})$ . By the same reasoning as above, each pair is divisible by $n$ . Now, if $n$ is even we end up with $\frac{n}{2}$ without a pair. Then, again by the factor of $2$ we have that this term is divisible by $n$ . So, $n\mid 2(1^k+2^k+\cdots +n^k)$ and so $n(n+1)\mid 2(1^k+2^k+\cdots +n^k)$ , as desired.

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