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.

Fernando Martin
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    http://math.stackexchange.com/questions/61798/generalisation-of-the-identity-sum-limits-k-1n-k3-bigg-sum-limits-k – jimjim Jun 27 '12 at 09:22
  • Look at this http://www.takayaiwamoto.com/Sums_and_Series/sumcube_1.html – pritam Jun 27 '12 at 09:29
  • See http://math.stackexchange.com/questions/120674 for remarks about proofs "not using induction". – sdcvvc Jun 27 '12 at 10:13
  • I merged the three existing posts which covered exactly this question, as each post had different interesting answers which should not be lost. I also deleted redundant comments, and comments about closing posts as duplicates. This [fourth question](http://math.stackexchange.com/questions/61798/generalisation-of-the-identity-sum-limits-k-1n-k3-bigg-sum-limits-k) is not considered a duplicate. – Eric Naslund Jul 02 '12 at 11:30
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    Since this question is asked frequently, it has been added to the [list of Generalizations of Common questions.](http://meta.math.stackexchange.com/questions/1868/list-of-generalizations-of-common-questions) It has been kept seperate from [the version which does use induction.](http://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct) – Eric Naslund Aug 30 '12 at 00:23
  • I know both formulae. I had never realized one was the square of the other...Thank you for asking the question. – Julien Apr 19 '13 at 20:40
  • Wikipedia calls this [Nicomachus's theorem](https://en.wikipedia.org/wiki/Nicomachus%27s_theorem) (from [Nicomachus of Gerasa](https://en.wikipedia.org/wiki/Nicomachus_of_Gerasa), also mentioned in some answers below). On Wikipedia, also see [Faulhaber's formula](https://en.wikipedia.org/wiki/Faulhaber%27s_formula). – Jeppe Stig Nielsen Oct 31 '16 at 13:05

27 Answers27


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

Proof that the sum of the cubes is the square of the sum

Parcly Taxel
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Mariano Suárez-Álvarez
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    original link: http://users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32 – Foo Bah Sep 03 '11 at 03:36
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    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 03 '11 at 04:13
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    I have put the details of the proof corresponding to this image in [this answer](http://math.stackexchange.com/questions/61482/intuitive-explanation-for-the-identity-sum-limits-k-1n-k3-left-sum-li/61544#61544) since the comment area was too small. – Wok Sep 03 '11 at 07:23
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    I don’t think this is a proof free from induction. – k.stm Mar 27 '14 at 12:58
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    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. – uniquesolution Oct 11 '15 at 00:39
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    That only shows you haven't stared at the image long enough... – Mariano Suárez-Álvarez Oct 11 '15 at 02:23
  • I had to cheat and look it up but once you see it you can't miss it. I'm ashamed of myself... – Karl Apr 01 '16 at 19:22

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

Graphic proof that the sum of cubes is the square of the sum of first powers

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

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  • 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 04 '11 at 07:39
  • 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 04 '11 at 07:49
  • I hope you don't mind if I use both ideas in [another answer](http://math.stackexchange.com/questions/61482/intuitive-explanation-for-the-identity-sum-limits-k-1n-k3-left-sum-li/61762#61762). – Wok Sep 04 '11 at 08:13
  • 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 04 '11 at 19:31

I believe this illustration is due to Anders Kaseorg:

Visual proof by block-stacking

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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.]

enter image description here enter image description here

The images are from Brian R Sears.


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

enter image description here

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 $1 + 2 + \ldots + n$.

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

enter image description here

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.

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

enter image description here

enter image description here

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: $$S_k^2 - S_{k-1}^2 = k \left(k+2 S_{k-1}\right) = k\left(k+k\left(k-1\right)\right)=k^3$$

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    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 04 '11 at 01:09
  • 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 04 '11 at 06:50
  • 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 04 '11 at 07:06

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

Bart Michels
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    But you're relying on a fact that is (often) proved with induction? –  Jun 26 '19 at 09:01
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    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?](https://math.stackexchange.com/a/120807/43288) – Bart Michels Jun 26 '19 at 09:09

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.

Qiaochu Yuan
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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
Martin Sleziak
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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.

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    why does the assumption hold? this is usually proved using induction... – akkkk Jun 27 '12 at 09:47
  • what assumption?? – Aang Jun 27 '12 at 09:48
  • I guess Auke means $\sum_{0}^{n} x^k=\frac{1-x^{n+1}}{1-x}$. – sdcvvc Jun 27 '12 at 10:12
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    LHS is a geometric series. http://en.wikipedia.org/wiki/Geometric_progression – Aang Jun 27 '12 at 10:14
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    @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 '12 at 12:40




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


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.

Thomas Andrews
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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 $n(n+1)$.

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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 $$B=\sum_1^n k=\frac{1}{2}n^2+\frac{1}{2}n$$ $A-B^2=0$. :)

Michael Hardy
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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)enter image description here $$7^2=\color{green}{1+2+3+4+5+6+7}\color{red}{+6+5+4+3+2+1}$$ Next, consider the standard multiplication table that we are all familiar with.The graphic shows the table up to the 9s.

enter image description here

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 $$\sum_{i=1}^n i^3 = \sum_{i=1}^n L_i = \sum_{i=1}^n R_i=\bigg(\sum_{i=1}^n i\bigg)^2$$

John Joy
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    Nice, you could rewrite that as an algebraic proof. It would be similar to my answer, but without subtracting $\sum k^2$. – Bart Michels Jul 06 '19 at 07:47

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

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


It seems to answer your question.

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

Fredrik Meyer
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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.

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

The trapezoids have area $k^3$ because their height equals $k$ and the $\text{width}_{\text{atHalfHeight}}$ consists of $k$ diagonals with width $k$: Trapezoidal layer with height = k and width = k^2

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

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

J. W. Tanner
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Job Bouwman
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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 $B_1'$.

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square triangular proof without words

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.

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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 $$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}=\Big(\frac{n(n+1)}{2}\Big)^2=\Big(\sum_{k=1}^nk\Big)^2$$

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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.

marty cohen
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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...

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To the collection of the geometric proofs:

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You've gotta see it to believe it.

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In formulas:

$$(\sum k)^2=\sum j^2 + 2 \sum_{i<j} ij =\sum_j (j^2+2 \sum_{i<j} ij)=\sum_j j(j+j(j-1))=\sum_j j(j^2)=\sum_j j^3$$

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 $j$.

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