5

$$(1+2+\cdots+n)^2=1^3+2^3+\cdots+n^3$$

I noticed this only because $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$ and $\displaystyle \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}$.

But the two things look completely different and I can't think of an intuitive reason I would have seen this connection. Something like the 'Proof Without Words' images. Or a trick I could use while expanding the right hand side to transform it to the sum of cubes.

genepeer
  • 1,678

2 Answers2

8

enter image description here

I think this image is due to Anders Kaseorg.

MJD
  • 65,394
  • 39
  • 298
  • 580
4

$$ \sum_{i=1}^n i^3 - \sum_{i=1}^{n-1} i^3 = n^3 $$

$$\begin{align} \left(\sum_{i=1}^n i\right)^2-\left(\sum_{i=1}^{n-1} i\right)^2&=\left(\sum_{i=1}^n i-\sum_{i=1}^{n-1} i\right)\left(\sum_{i=1}^n i+\sum_{i=1}^{n-1} i\right)\\ &=n\left(\sum_{i=1}^n i + \sum_{i=1}^{n-1} (n-i)\right)\\ &=n\left(n + \sum_{i=1}^{n-1} n\right)\\ &=n\left(n + n(n-1)\right)\\ &=n\cdot n^2 = n^3 \end{align}$$

(This is to show that it makes mathematically intuitive sense)

Glen O
  • 12,425
  • Nice. I wasn't sure what you were doing till I saw the last sentence – genepeer Apr 25 '13 at 02:15
  • It's a very nice and tricky answer, mostly because you did not use the formula $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$ in your induction. – Woria Oct 07 '17 at 17:24