Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction

I recently proved that

$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$

using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.

28 Answers

Stare at the following image, taken from this MO answer, long enough:

• original link: users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32 Sep 3, 2011 at 3:36
• The fact that there are $k$ blocks (or $\frac{1}{2}+k{-}1+\frac{1}{2}$ blocks) of $k\times k$ size is based on the fact that $\sum\limits_{j=1}^{k-1}=k(k{-}1)/2$. That is, $(k{-}1)/2$ blocks on top $(k{-}1)/2$ on the left and $1$ block at the corner (totaling to $k$). Perhaps I am being picky or slow, but I don't see this as obvious from the image. Beyond that, it is a nice proof-without-words.
– robjohn
Sep 3, 2011 at 4:13
• I don’t think this is a proof free from induction. Mar 27, 2014 at 12:58
• This only proves the assertion for $n=5$. If someone is to accept its validity for general $n$, then it is clearly not induction-free. Oct 11, 2015 at 0:39
• That only shows you haven't stared at the image long enough... Oct 11, 2015 at 2:23

I don't know if this is intuitive, but it is graphic.

On the outer edge of each $(k{+}1){\times}k$ block there are $k$ pairs of products each of which total to $k^2$. Thus, the outer edge sums to $k^3$, and the sum of the whole array is therefore $\sum\limits_{k=1}^n k^3$.

The array is the matrix product $$\left[\begin{array}{r}0\\1\\2\\\vdots\\n\end{array}\right]\bullet\left[\begin{array}{rrrrr}1&2&3&\cdots&n\end{array}\right]$$ Therefore, the sum of the elements of the array is $\sum\limits_{k=0}^nk\;\sum\limits_{k=1}^nk=\left(\sum\limits_{k=1}^nk\right)^2$.

Therefore, $\sum\limits_{k=1}^n k^3=\left(\sum\limits_{k=1}^nk\right)^2$

• If we forget the first line of the matrix (which is zero and which is only used to make pairs with the diagonal coefficients), then I like the fact we can put this answer in parallel with the coloured rectangles above and below, and we get another partition of each colored area (each coefficient of the matrix gives a rectangle of a certain area), which explains why each L-shaped area is $k^3$.
– Wok
Sep 4, 2011 at 7:39
• However, each L-shape coefficient has the same factor $k$, which means it proves "each L-shaped area is $k^3$" by the same proof that $2 \times \sum_{j=0}^k j = (0+k) + (1+k-1) + ... + (k-1+1) + (k+0) = (k+1) \times k$, which makes it really close to the coloured rectangles above and below.
– Wok
Sep 4, 2011 at 7:49
• I hope you don't mind if I use both ideas in another answer.
– Wok
Sep 4, 2011 at 8:13
• I don't mind. I simply find it less aesthetic to need to use $\sum\limits_{j=1}^k\;j=k(k+1)/2$ or that $(k(k+1)/2)^2-(k(k-1)/2)^2=k^3$ in an intuitive proof.
– robjohn
Sep 4, 2011 at 19:31

I believe this illustration is due to Anders Kaseorg:

Can you get the intuition explanation from the following two pictures?[EDIT: the following is essentially the same as Mariano's answer. He didn't mentioned the first picture though.]

The images are from Brian R Sears.

• nice, but already posted (linked) by Mariano Sep 2, 2011 at 23:24

Here's another version of this "proof without words". This is the case $n=4$.

There are 1 $1 \times 1$, 2 $2 \times 2$, 3 $3 \times 3$, ... squares, for a total area of $1^3 + 2^3 + \ldots + n^3$. For even $k$, two of the $k \times k$ squares overlap in a $k/2 \times k/2$ square, but this just balances out a $k/2 \times k/2$ square that is left out, so the total is the area of a square of side $1 + 2 + \ldots + n$.

There's this nice picture from the Wikipedia entry on the squared triangular number:

The left side shows that $1 + 2 + 3$ forms a triangle and so that squaring it produces a larger triangle made up of $1+2+3$ copies of the original triangle. The right side has $1(1^2) + 2(2^2) + 3(3^2) = 1^3 + 2^3 + 3^3$. The coloring shows why the two sides are equal.

There are several other references for combinatorial proofs and geometric arguments on the Wikipedia page.

Here's a direct algebraic proof. $$\sum_{k=1}^n(k^3-k^2)=2\sum_{k=1}^nk\cdot\frac{k(k-1)}2=2\sum_{k=1}^nk\sum_{l=1}^{k-1}l=2\sum_{1\leqslant l

Here's also a nice analogue of this calculation:

$$\int_0^x t^3dt = 2 \int_0^x t \cdot \frac{t^2}2dt = 2\int_0^x t dt\int_0^t udu = \left(\int_0^x t\right)^2 \,.$$

Just to emphasize that the proof relies on the substitution rule, Fubini's theorem, and the discrete version of $$\int_0^x t dt= \frac{x^2}2$$, but not on induction directly.

• But you're relying on a fact that is (often) proved with induction?
– user636532
Jun 26, 2019 at 9:01
• Definitely. But one that could credibly pretend not to rely on induction. I very much like the last paragraph of this answer: Must we use induction to prove a statement for all integers? Jun 26, 2019 at 9:09
• Jul 4, 2022 at 11:29

Each colored area is $k^3$ as a difference of two areas: $S_k^2 - S_{k-1}^2$.

The detailed proof which comes with the drawing is the following.

For any positive integer $k$, we define: $$S_i = \sum_{j=1}^{i} j$$

We first notice: $$S_i^2 = S_i^2 - S_0^2= \sum_{k=1}^{i} \left(S_k^2 - S_{k-1}^2\right)$$

The expected result finally comes from: $$S_k^2 - S_{k-1}^2 = k \left(k+2 S_{k-1}\right) = k\left(k+k\left(k-1\right)\right)=k^3$$

• So essentially, you are using the fact that $$\left(\sum_{j=1}^k\;j\right)^2-\left(\sum_{j=1}^{k-1}\;j\right)^2=k^3$$ to justify the diagram which is supposed to prove that fact intuitively.
– robjohn
Sep 4, 2011 at 1:09
• As you mentioned in another answer, this is where the diagram is the least intuitive. However, if you cannot picture $k^3$ on a 2D-plane, then you need another representation as a difference of areas.
– Wok
Sep 4, 2011 at 6:50
• As soon as you know $S_k = \sum_{j=1}^k j = \frac{k(k+1)}{2}$, I find it more intuitive to figure out each colored area is $k^3$ on this diagram than to figure it out by counting squares plus two rectangles when $k$ is even.
– Wok
Sep 4, 2011 at 7:06

The formula is due to Nicomachus of Gerasa. There is a nice discussion of ways to prove it at this n-category cafe post, including a bijective proof and some visual / "geometric" proofs.

Several visual proofs of this indentity are collected in the book Roger B. Nelsen: Proofs without Words starting from p.84.

Although several of these proofs can still be considered inductive, I thought it might be interesting to mention them.

Original sources are given on p. 147:

• 84 Mathematical Gazette, vol. 49, no. 368 (May 1965), p. 199. jstor
• 85 Mathematics Magazine, vol. 50, no. 2 (March 1977), p. 74. jstor
• 86 Mathematics Magazine, vol. 58, no. 1 (Jan. 1985), p. 11. jstor
• 87 Mathematics Magazine, vol. 62, no. 4 (Oct. 1989), p. 259. jstor
• 87 Mathematical Gazette, vol. 49, no. 368 (May 1965), p. 200. jstor
• 88 Mathematics Magazine, vol. 63, no. 3 (June 1990), p. 178. jstor
• 89 Mathematics Magazine, vol. 62, no. 5 (Dec. 1989), p. 323. jstor
• 90 Mathematics Magazine, vol. 65, no. 3 (June 1992), p. 185. jstor
• Great collections. Thanks Jun 27, 2012 at 11:11

You know, $$\sum_0^n x^k=\frac{1-x^{n+1}}{1-x}$$. Differentiate both sides once, $$\sum_1^n kx^{k-1}=\frac{x^n(nx-n-1)+1}{x^2-2x+1}$$. Now taking $$\lim_{x\to1}$$ both sides and then squaring the result will give you the expression on the RHS. You can further differentiate $$\sum_0^n x^k=\frac{1-x^{n+1}}{1-x}$$ until you get $$k^3$$ inside the expression, take limit again you will get the same result as of $$\left(\lim_{x\to1}\frac{x^n(nx-n-1)+1}{x^2-2x+1}\right)^2$$. You can also prove it using telescopic series.

• why does the assumption hold? this is usually proved using induction... Jun 27, 2012 at 9:47
• what assumption??
– Aang
Jun 27, 2012 at 9:48
• I guess Auke means $\sum_{0}^{n} x^k=\frac{1-x^{n+1}}{1-x}$. Jun 27, 2012 at 10:12
• LHS is a geometric series. en.wikipedia.org/wiki/Geometric_progression
– Aang
Jun 27, 2012 at 10:14
• @Auke: one can also prove it like this - let $f(x) = \sum\limits_{k=0}^nx^k$, then $f(x) - xf(x)$ is: \begin{align} 1+&x+x^2+\dots+x^n \\ &- \\ &x+x^2+\dots+x^n+x^{n+1} \end{align} which is $1-x^{n+1}$. Hence $$(1-x)f(x) = 1-x^{n+1}$$ as needed.
– Ilya
Jun 27, 2012 at 12:40

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

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

$$f(n)-f(n-1)=n^3$$

if $$g(n)= (1+2+3+4+\cdots+n)^2$$ then

$$g(n)-g(n-1)=(1+2+3+4+\cdots+n)^2-(1+2+3+4+\cdots+(n-1))^2$$

using $$a^2-b^2=(a+b)(a-b)$$

\begin{align}g(n)-g(n-1)&=\\ &=[(1+\dots+n)+(1+\dots +(n-1))][(1+\dots+ n)-(1+\dots+(n-1)]\\ &=[2(1+2+3+4+\cdots+(n-1))+n]n\\ &=\left(2\frac{n(n-1)}{2}+n\right)n\\ &==(n(n-1)+n)n\\ &=n^3 \end{align}

So $$f(n)$$ and $$g(n)$$ are equal.

The sum of a degree $n$ polynomial $f(n)$ will be a degree $n+1$ polynomial $S(n)$ for $n \geq 0$ and both polynomials can be extended (maintaining the relation $S(n)-S(n-1) = f(n)$) to negative $n$. To verify that the formula for $\Sigma k^3$ is correct one need only test it for any 5 distinct values of $n$, but the structure of the answer can be predicted algebraically using the continuation to negative $n$.

If $S(n) = (1^3 + 2^3 + \dots n^3)$ is the polynomial that satisfies $S(n)-S(n-1) = n^3$ and $S(1)=1$, then one can calculate from that equation that $S(0)=S(-1)=0$ and $S(-n-1)=S(n)$ for all negative $n$, so that $S$ is symmetric around $-1/2$. The vanishing at 0 and -1 implies that $S(t)$ is divisible as a polynomial by $t(t+1)$. The symmetry implies that $S(t)$ is a function (necessarily a polynomial) of $t(t+1)$.

$S(t)$ being of degree 4, this means $S(n) = a (n)(n+1) + b((n^2 +n)^2$ for constants $a$ and $b$. Summation being analogous to integration (and equal to it in a suitable limit), they have to agree on highest degree terms. Here it forces $b$ to be $1/4$ to match $\int x^3 = x^4/4$. Computing the sum at a single point such as $n=1$ determines $a$, which is zero.

Similar reasoning shows that $S_k(n)$ is divisible as a polynomial by $n(n+1)$ for all $k$. For odd $k$, $S_k(n)$ is a polynomial in $n(n+1)$.

We know that $$A=\sum_1^n k^3=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2$$ and $$B=\sum_1^n k=\frac{1}{2}n^2+\frac{1}{2}n$$ $A-B^2=0$. :)

• If you are presenting this proof,write it nicely. $A=\frac{n^2(n+1)^2}{4}$ and $B=\frac{n(n+1)}{2}$,obviously $A=B^2$
– Aang
Jun 27, 2012 at 10:40
• @avatar: Dear Avatar; I will. Your proof is great and it looks analytically. Thanks. :-) Jun 27, 2012 at 10:48
• sorry, if you felt that i was being rude.
– Aang
Jun 27, 2012 at 10:53
• @avatar: Your proof lights mine. No Problem at all. WELCOME Avatar. :) :) Jun 27, 2012 at 10:57

I find a quite funny proof!

Observe that $$n^3=n^2\times n$$ By using the identity $$xy=\left(\dfrac{x+y}{2}\right)^2-\left(\dfrac{x-y}{2}\right)^2$$, we get $$n^2\times n=\left(\dfrac{n^2+n}{2}\right)^2-\left(\dfrac{n^2-n}{2}\right)^2=\left(\sum_{k=0}^n k\right)^2-\left(\sum_{k=0}^{n-1} k\right)^2$$ Therefore, $$\sum_{r=1}^n r^3=\sum_{r=1}^n \left[\left(\sum_{k=0}^r k\right)^2-\left(\sum_{k=0}^{r-1} k\right)^2\right]=\left(\sum_{k=0}^n k\right)^2-\left(\sum_{k=0}^{0} k\right)^2=\left(\sum_{k=1}^n k\right)^2$$ After changing the variables, we get $$\boxed{\sum_{k=1}^n k^3=\left(\sum_{k=1}^n k\right)^2}$$

• VERY VERY VERY NICE!!!!! This is a wonderfully elegant proof (+1) Dec 30, 2019 at 8:16

One way to show that $$\sum_{i=1}^n i^3 = \bigg(\sum_{i=1}^n i \bigg)^2$$ is to add up the entries in the multiplication tables, but first we need to show that $$1+2+3+\dots+n+\dots+3+2+1 = n^2$$ For this, see the image below (n=7) $$7^2=\color{green}{1+2+3+4+5+6+7}\color{red}{+6+5+4+3+2+1}$$ Next, consider the standard multiplication table that we are all familiar with.The graphic shows the table up to the 9s.

We can add up the entries in any order that we wish. One way would be to add up a series of Ls (the 6th L ($$L_6$$) is highlighted in yellow). \begin{align} L_6 &= 6+12+18+24+30+36+30+24+18+12+6\\ &=6(1+2+3+4+5+6+5+4+3+2+1)\\ &=6(6^2)\\ &=6^3 \end{align} And the sum of all the entries in the table becomes $$\sum_{i=1}^n L_i = \sum_{i=1}^n i^3$$ Alternatively, we could just add up each row. The 6th row ($$R_6$$) would be \begin{align} R_6 &= 6+12+18+24+30+36+42+48+54\\ &= 6(1+2+3+4+5+6+7+8+9)\\ &= 6\sum_{i=1}^9 i \end{align} And the sum of all the entries becomes $$\sum_{i=1}^n R_i = \sum_{i=1}^n \bigg(i\sum_{j=1}^n j\bigg)=\bigg(\sum_{j=1}^n j\bigg)\bigg(\sum_{i=1}^n i\bigg)=\bigg(\sum_{i=1}^n i\bigg)^2$$ Thus we have $$\sum_{i=1}^n i^3 = \sum_{i=1}^n L_i = \sum_{i=1}^n R_i=\bigg(\sum_{i=1}^n i\bigg)^2$$

• Nice, you could rewrite that as an algebraic proof. It would be similar to my answer, but without subtracting $\sum k^2$. Jul 6, 2019 at 7:47

Chance would have it that I stumbled* upon this article today:

http://blogs.mathworks.com/loren/2010/03/04/nichomachuss-theorem/

It seems to answer your question.

(* That is, @AlgebraFact on Twitter posted a link)

http://en.wikipedia.org/wiki/Faulhaber%27s_formula#Faulhaber_polynomials

If $p$ is odd, then $1^p+2^p+3^p+\cdots+n^p$ is a polynomial function of $a=1+2+3+\cdots+n$. If $p=3$, then then the sum is $a^2$; if $p=5$ then it's $(4a^3-a^2)/3$, and so on.

• And $a$ is always a factor of $p$.
– lhf
Sep 3, 2011 at 13:26

Here's a simple bijective proof of a different sort:

Consider a staircase with $n$ steps, built out of $\sum_{k=1}^n k$ blocks. In other words, take the set $\{(i,j) \in \mathbb{Z}\times\mathbb{Z}: i + j \leq n, i > 0, j > 0\}$.

Then $\left(\sum_{k=1}^n k\right)^2$ is the number of ordered pairs $(B_1,B_2)$ of blocks.

And $\sum_{k=1}^n k^3$ is the number of ordered $4$-tuples $(k,b_1,b_2,b_3)$, where $k \in \{1,\ldots,n\}$, and $b_1$ is along the $k$-th diagonal $b_1 \in \{(k+1-j,j): j \in \{1,\ldots,k\}\}$, and $b_2$ is along the bottom $b_2 \in \{(j,1): j \in \{1,\ldots, k\}\}$ and $b_3$ is along the left side $b_3 \in \{(1,j): j \in \{1, \ldots, k\}\}$.

The bijection:

Given an ordered tuple $(k,b_1,b_2,b_3)$, let $A_1 = b_1$ and let $A_2$ given by $b_2$ and $b_3$ as its $x$ and $y$ coordinates, so if $b_2 = (i,1)$ and $b_3 = (1,j)$ then $A_2 = (i,j)$.

Case 1: $A_2$ is on or below the $k$-th diagonal. Then let $(B_1, B_2) = (A_1, A_2)$.

Case 2: $A_2$ is above the $k$-the diagonal. Then let $A_2'$ be the reflection across the $k$-th diagonal of $A_2$. That is, if $A_2 = (i,j)$ then $A_2' = (k+1-j,k+1-i)$. Then let $(B_1, B_2) = (A_2', A_1)$.

The inverse:

To get the inverse, take whichever of $B_1$ and $B_2$ is on a higher diagonal (i.e. has greater sum of its coordinates), taking $B_1$ in case of ties, and let that be $b_1$ and let $k$ be the sum of the coordinates of $b_1$.

Case 1: If $B_1$ is used: Take $B_2$ and let $b_2$ and $b_3$ be given by points with the same the $x$- and $y$-coordinates, respectively, as $B_2$.

Case 2: If $B_2$ is used: Take $B_1'$ (i.e. the reflection across the $k$-th diagonal, as above) and let $b_2$ and $b_3$ be given by points with the same the $x$- and $y$-coordinates, respectively, as $B_1'$.

The square in the identity is the area of the triangle below, while the cubes are the areas of the trapezoidal layers, with heights $$k = 1, 2, \cdots, n$$

The trapezoids have area $$k^3$$ because their height equals $$k$$ and the $$\text{width}_{\text{atHalfHeight}}$$ consists of $$k$$ diagonals with width $$k$$:

The total of the triangle is its squared height $$(1 + 2 + \cdots + n)^2$$, because this triangle can be turned into a square:

Therefore: $$(1 + 2 + \cdots + n)^2 = \sum_{k=1}^n k^3$$ , $$\blacksquare$$

• Is there some mathematical explanation, why the triangle can be turned into square? Sep 6, 2020 at 5:28

This is about the same proof as here, the presentation is a bit different though. This is another way to make $k^3$ appear than what was shown here, here and here.

For every $k\in\mathbb{N}$ $$(k+1)^4=k^4+4k^3++6k^2+4k+1$$ therefore $$\sum_{k=1}^n(k+1)^4=\sum_{k=1}^nk^4+4\sum_{k=1}^nk^3+6\sum_{k=1}^nk^2+4\sum_{k=1}^nk+\sum_{k=1}^n1$$ which is equivalent to $$\sum_{k=1}^nk^4+(n+1)^4-1=\sum_{k=1}^nk^4+4\sum_{k=1}^nk^3+6\sum_{k=1}^nk^2+2n(n+1)+n$$ After simplifications we obtain $$4\sum_{k=1}^nk^3=(n+1)^4-1-2n(n+1)-n-6\sum_{k=1}^nk^2=n^4+4n^3+4n^2+n-6\sum_{k=1}^nk^2$$ Using $$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}\hspace{0.2cm}\text{and}\hspace{0.2cm}\sum_{k=1}^nk=\frac{n(n+1)}{2}$$ we get $$4\sum_{k=1}^nk^3=n^4+4n^3+4n^2+n-6\sum_{k=1}^nk^2\\=n^4+4n^3+4n^2+n-n(n+1)(2n+1)\\=n^4+2n^3+n^2=n^2(n+1)^2$$ Finally $$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}=\Big(\frac{n(n+1)}{2}\Big)^2=\Big(\sum_{k=1}^nk\Big)^2$$

We begin by writing $$k^3$$ in a more clever fashion: $$k^3 = k(k-1)(k-2) + 3k^2 - 2k$$ :

$$\sum_{k=0}^n k^3 = \sum_{k=0}^n k(k-1)(k-2) + 3k^2 -2k$$

Distributing the summation and adjusting our indices we obtain: $$\sum_{k=3}^n k(k-1)(k-2) + \sum_{k=0}^n 3k^2 - \sum_{k=0}^n 2k$$

Notice, $$k(k-1)(k-2) = \frac {k!}{(k-3)!}$$ Now we have $$\sum_{k=3}^n \frac {k!}{(k-3)!} + \sum_{k=0}^n 3k^2 - \sum_{k=0}^n 2k$$ Notice that we have nearly obtained the binomial expansion of K choose 3, all we need to do is divide by 3! So we offset this by also taking the product of 3! $$\sum_{k=3}^n \frac {k!}{(k-3)!} = 3!\sum_{k=3}^n \frac {k!}{(k-3)!3!} = 3!\sum_{k=3}^n\binom{k}{3} = 3!\binom{n+1}{4}$$ We have now obtained $$\sum_{k=0}^n k^3 = 3!\binom{n+1}{4} + 3\sum_{k=0}^n k^2 - 2\sum_{k=0}^n k$$ Focusing solely on the right-hand side we have $$6\biggl(\frac {(n+1)!}{(n-3)!4!}\biggr) + 3\sum_{k=0}^n k^2 - 2\sum_{k=0}^n k$$ Assuming we already know the sum of the sequence of integers and squared integers (the 2 sums we have left) we have $$\frac {(n+1)(n)(n-1)(n-2)}{4} + 3\frac{n(n+1)(2n+1)}{6} - 2\frac {n(n+1)}{2}$$ Generating common denominators and with a bit of algebra we now have $$\frac {n^4-2n^3-n^2+2n+4n^3+6n^2+2n-4n^2-4n}{4}$$ Combining like-terms we have reached our solution: $$\frac {n^4+2n^3+n^2}{4} = \biggl(\sum_{k=0}^n k\biggr)^2$$

I was just playing around, and I am probably the 47,000,000th person to do this, and this is undoubtedly equivalent to another answer but, anyway...

Use $$\sum_{i=1}^n i =i(i+1)/2$$ and $$(\sum_{i=1}^n a_i)^2 =\sum_{i=1}^n a_i^2 +2\sum_{i=1}^n\sum_{j=1}^{i-1} a_ia_j$$.

$$\begin{array}\\ (\sum_{i=1}^n i)^2 &=\sum_{i=1}^n i^2 +2\sum_{i=1}^n\sum_{j=1}^{i-1} ij\\ &=\sum_{i=1}^n i^2 +2\sum_{i=1}^ni\sum_{j=1}^{i-1} j\\ &=\sum_{i=1}^n i^2 +2\sum_{i=1}^ni\cdot i(i-1)/2\\ &=\sum_{i=1}^n i^2 +\sum_{i=1}^ni^2(i-1)\\ &=\sum_{i=1}^n i^2 +\sum_{i=1}^n(i^3-i^2)\\ &=\sum_{i=1}^n i^2 +\sum_{i=1}^ni^3-\sum_{i=1}^ni^2\\ &=\sum_{i=1}^ni^3\\ \end{array}$$

It was a pleasant surprise that $$\sum_{i=1}^ni^2$$ cancelled out.

After many years I still think the best way to solve this kind of problem in a natural and systematic way is to view it as a recurrence relation with constants coefficients, in this case, $x_n = x_{n-1}+n^3$. The way I learnt to do so was by using characteristic polynomial but you may prefer some other method...

To the collection of the geometric proofs:

You've gotta see it to believe it.

In formulas:

$$(\sum k)^2=\sum j^2 + 2 \sum_{i

In words: we can assemble the square with side $$(1+...+k)$$ from $$k$$ smaller squares and pairs of rectangular blocks. If we look only at the rectangular blocks with the largest side fixed to some size $$j$$, they all pair up into $$j-1$$ "complementary" pairs (based on the size of the smaller side, pairing $$1$$ with $$j-1$$, $$2$$ with $$j-2$$, untill $$j-1$$ with $$j$$; this is Gauss' trick), each pair of rectangles forming a $$j$$ by $$j$$ square together. Thus, together with the $$j^2$$ piece they assemble into a cube of side $$j$$.

This answer uses the same sort of reasoning as given in the answer by zyx. I show that we can get a bit more out of the recursion relation satisfied by the function $$S(n)$$ and their generalizations for summation of arbitrary powers, in particular that $$S(t)$$ (and its generalization for summations over odd powers larger than 1) contains a factor $$t^2 (1+t)^2$$. This then immediately establishes the result.

We start with defining the functions $$S_k(n)$$:

$$S_k(n) =\sum_{r=0}^{n}r^k \tag{1}$$

where $$k$$ and $$n$$ are integers and we take $$k\geq 1$$ and $$n\geq 0$$. The functions $$S_k(n)$$ satisfy the recursion:

$$S_k(n) - S_k(n-1) = n^k \tag{2}$$

This recursion together with the condition that $$S_k(0) = 0$$ which follows from (1), fixes the function $$S_k(n)$$. In particular, it follows from this that $$S_k(n)$$ is a polynomial of degree $$k+1$$ in $$n$$. This means that we can lift the constraint that $$n$$ be an integer larger than or equal to zero and instead take $$n$$ to be any arbitrary real or complex number.

If we put $$n = 0$$ in (2) and use that $$S_k(0) = 0$$, we find:

$$S_k(-1) = 0$$

The function $$S_k(x)$$ is therefore a polynomial of degree $$k+1$$ which contains a factor of $$x (x+1)$$. Since $$S_1(x)$$ is a second degree polynomial, this fixes $$S_1(x)$$ up to a constant factor, and we can find that this factor is $$\frac{1}{2}$$ by using $$S_k(1) = 1$$, although that's not necessary to do for the purpose of this problem.

We can find a symmetry relation satisfied by $$S_k(n)$$ using the recursion (2) as follows. We replace $$n$$ by $$-n$$ to obtain:

$$S_k(-n) - S_k(-n-1) = (-1)^k n^k$$

Let’s define the function $$\tilde{S}_{k}(n) = S_k(-n)$$. We can then write the above recursion as:

$$\tilde{S}_k(n+1) - \tilde{S}_k(n) = (-1)^{k+1} n^k$$

We then see that the function $$(-1)^{k+1}\tilde{S}_k(n+1)$$ satisfies the same recursion as the function $$S_k(n)$$. The value at $$n = 0$$ is equal to zero, because $$\tilde{S}_k(1)= S_k(-1) = 0$$. This means that these two functions are identical. We thus have:

$$S_k(-x-1) = (-1)^{k+1}S_k(x)$$

The point at which $$x = -x - 1$$ is $$x = -\frac{1}{2}$$. We see that for even $$k$$, $$S_k(x)$$ is anti-symmetric w.r.t. reflection about this point, while in case of odd $$k$$, it is symmetric. This means that for even $$k$$, the function has a zero there, while for odd $$k$$ the derivative is zero:

$$\begin{split}S_{2k}\left(-\frac{1}{2}\right) &= 0\\ S'_{2k+1}\left(-\frac{1}{2}\right) &= 0\end{split}$$

This fixes $$S_2(x)$$, because $$S_2(x)$$ is a third degree polynomial and it has zeros at $$x = 0$$, $$x = 1$$ and $$x = -\frac{1}{2}$$, so it’s proportional to $$x (x+1)(2x+1)$$, and demanding that $$S_2(1) = 1$$ yields $$S_2(x) = \frac{1}{6} x (x+1)(2x+1)$$. Obtaining this expression is, however, not necessary for the purpose of this problem.

In general, we have that $$S_{2k}(x)$$ contains a factor of $$x(x+1)(2x+1)$$. We can obtain more information about $$S_k(x)$$ for odd $$k$$ by taking the derivative of the recursion (2). This yields:

$$S'_{2k+1}(x) - S'_{2k+1}(x-1) = (2k+1) x^{2k}$$

We then see $$\frac{S'_{2k+1}(x)}{2k+1}$$ satisfies the same recursion as $$S_{2k}(x)$$. Because both these functions are zero at $$x = -\frac{1}{2}$$, they are identical. The fact that $$S_{2k}(x)$$ is zero at $$x = 0$$ and $$x = 1$$ then tells us that $$S'_{2k+1}(x)$$ is zero there as well. So, we see $$S_{2k+1}(x)$$ has zeros at $$x = 0$$ and $$x = 1$$ of multiplicity of at least $$2$$, so $$S_{2k+1}(x)$$ contains a factor of $$x^2 (x+1)^2$$.

In particular, since $$S_3(x)$$ is a fourth degree polynomial, we see that it is proportional to $$x^2(x+1)^2$$, so it's proportional to $$S_1(x)^2$$. And since $$S_k(1) =1$$for all $$k$$, we have $$S_3(x) = S_1(x)^2$$.