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I am trying to prove the following summation:
$$\sum_{n = 1}^{2160}\exp\left(i\pi\frac{n(n + 1)(2n + 1)}{1080}\right) = 0$$ I suspect that some periodicity or symmetry in the expression plays a key role in simplifying the proof, but I am not sure how to proceed rigorously.

I would appreciate a detailed explanation or guidance on how to approach this problem. Specifically:

  • How can periodicity or modular arithmetic simplify the summation?
  • Are there properties of $e^{i\pi x}$ or symmetry arguments that lead to the result?

Any insight or suggestions would be very helpful. Thank you!

CrSb0001
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Faust42
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    Hint: $\frac{n(n+1)(2n+1)}6$ is the formula for the sum of squares. This gives that the sum is equal to $$\sum_{n=1}^{2160}e^{i\pi\cdot(1/180)\cdot\sum_{i=1}^ni^2}$$And now given 2160/180=12, can you take it from here? – CrSb0001 Dec 18 '24 at 15:36
  • Hint: $1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}6$ – User Dec 18 '24 at 15:36
  • @CrSb0001 $2160/180=12$... so? – Davide Masi Dec 18 '24 at 20:29
  • From a computation, the sum is zero because every term appears as $t$ and $-t$. – lhf Dec 18 '24 at 23:01

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