# Formalize a proof without words of the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$

This website gives the following proof without words for the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$. I find it interesting but have trouble seeing the proof behind it. Could anyone could give me a formal proof based on this animation?

• so you are asking for the proof of 40.media.tumblr.com/18e75c6163b5b9855306e91703d7f1a9/… ? by induction! – Mister Benjamin Dover Jan 3 '15 at 19:38
• @Laters yes! That's what I figure. I'm just being lazy because I'm on college break. – mconn7 Jan 3 '15 at 19:39
• I think the question is not asking for just a proof (if so, induction is the obvious approach), but rather for a particular proof which formalizes the construction used in the gif. – 6005 Jan 3 '15 at 19:43
• see this also the first comment under the question – Myself Jan 3 '15 at 19:48
• This is a terrible way to present a question. The title ("Can you should me a formal proof of this gif?") gives no hint as to what the mathematical question is, neither does the body of the question, and the tag, "proof-strategy", is no help at all. Too lazy to search for a duplicate, and too uncaring to provide a title that would help future users search for a duplicate. Please don't misuse this site this way. – Gerry Myerson Jan 3 '15 at 19:55

## 2 Answers

This gif is a nice visualization of the following algebraic manipulations (color to correspond to the colors in the gif): \begin{align*} (1 + 2 + 3 + \cdots + n)^2 &= \sum_{i=1}^n \sum_{j=1}^n ij \\ &= \color{green}{\sum_{1 \le i < j \le n} ij} + \color{red}{\sum_{1 \le j < i \le n} ji} + \color{blue}{\sum_{1 \le j \le n} j^2} \\ &= \color{green}{\sum_{1 \le i < j \le n} ij} + \color{red}{\sum_{1 \le i < j \le n} ji} + \color{blue}{\sum_{1 \le j \le n} j^2} \\ &= \sum_{j=1}^n j \left[ \color{green}{\sum_{i=1}^{j-1} i} + \color{red}{\sum_{i=1}^{j-1} i} + \color{blue}{j} \right] \\ &= \sum_{j=1}^n j \left[ \sum_{i=1}^{j-1} \big(\color{green}{(i)} + \color{red}{(j-i)}\big) + \color{blue}{j} \right] \\ &= \sum_{j=1}^n j \left[ \sum_{i=1}^j j\right] \\ &= \sum_{j=1}^n j^3. \end{align*} Notice that the base of the $n^\text{th}$ cube is the diagonal square prism $n^2\times 1$, and that each cube is build out of the diagonal square prism and rectangular prisms above and to the left of the diagonal square prism. Since adding another term to the left hand sum will give us one more diagonal square and set of prisms above and left of it, this suggests we can use induction.

For $n=1$, we trivially have $\left(\sum_{m=1}^n m\right)^2 = \sum_{m=1}^n m^3$.

Now, let's suppose $\left(\sum_{m=1}^n m\right)^2 = \sum_{m=1}^n m^3$ for some $n$. Then, \begin{align*}\left(\sum_{m=1}^{n+1} m\right)^2 &= \left(n+1 + \sum_{m=1}^{n} m\right)^2\\ &= (n+1)^2 + 2(n+1)\left(\sum_{m=1}^{n} m\right) + \left(\sum_{m=1}^{n} m\right)^2 \tag{*}\\ &= (n+1)^2 + 2(n+1)\frac{n(n+1)}{2} + \sum_{m=1}^{n} m^3 \tag{**}\\ &= (n+1)^2(n+1) + \sum_{m=1}^{n} m^3\\ &= \sum_{m=1}^{n+1} m^3 \end{align*} where $(**)$ comes from the well known identity $\sum_{m=1}^n m = \frac{n(n+1)}{2}$. In $(*)$, the first term is the volume of the diagonal square prism, the middle term is the volume of the rectangular prisms left of and above the diagonal prism, and the last term is the volume of the remaining shapes.

• @KristofferRyhl Thanks, fixed! – Strants Jan 3 '15 at 19:56
• Thank you! That makes so much sense! @Strants – mconn7 Jan 3 '15 at 20:06