# Intuitive/Visual proof that $(1+2+\cdots+n)^2=1^3+2^3+\cdots+n^3$ [duplicate]

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

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.

• There are two pictures on this Wiki article which might be of interest. Apr 25, 2013 at 2:02
• Duplicate of math.stackexchange.com/q/61482/264 Apr 25, 2013 at 2:04
• I would delete the question but we'd lose the image posted by MJD! Apr 25, 2013 at 2:17
• I reposted it at the other place too.
– MJD
Apr 25, 2013 at 2:22
• @genepeer: MJD has posted his answer on the other thread now. There's no need to delete your question; we'll just close it. Apr 25, 2013 at 2:22 I think this image is due to Anders Kaseorg.

• Very pretty! How did you make the picture? Apr 25, 2013 at 2:06
• I didn't make it. I did a search on this site for "sum of cubes", found this question, and plucked the image out of a deleted post.
– MJD
Apr 25, 2013 at 2:08
• This is awesome! Apr 25, 2013 at 2:12

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

\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)

• Nice. I wasn't sure what you were doing till I saw the last sentence Apr 25, 2013 at 2:15
• 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. Oct 7, 2017 at 17:24