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I recently brushed up on the formulas

\begin{align} \sum_{i = 1}^n i &= \frac{n(n+1)}{2},\\ \sum_{i = 1}^n i^2 &= \frac{n(2n+1)(n+1)}{6},\\ \sum_{i = 1}^n i^3 &= \frac{n^2(n+1)^2}{4}, \end{align}

and I could not believe I never noticed the identity $\sum_{i = 1}^n i^3 = \left(\sum_{i = 1}^n i\right)^2$. Is there an obvious reason why this should be the case?

I purposefully kept the question open to interpretation because I would love to see how you reason about this fact that I should have noticed ages ago.

kjQtte
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  • +1, good question. Obviously, it should be easy to prove, like via induction. But if the question is, "Is it obvious?" Then no, not to me... – Adam Rubinson Aug 11 '23 at 19:53
  • @AdamRubinson I changed the question, thank you. – kjQtte Aug 11 '23 at 20:00
  • See https://math.stackexchange.com/q/61482/96384, https://math.stackexchange.com/q/1882161/96384, and many other duplicates. – Torsten Schoeneberg Aug 11 '23 at 20:01
  • Yes it does, I wasn't able to find those. Thank you. I'll keep the post up so it's easier to find for someone else if they ever have the same question. – kjQtte Aug 11 '23 at 20:05

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