# 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.

Stare at the following image, taken from this MO answer, long enough: • original link: users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32 – Foo Bah Sep 3 '11 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 '11 at 4:13
• I don’t think this is a proof free from induction. – k.stm Mar 27 '14 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. – uniquesolution Oct 11 '15 at 0:39
• That only shows you haven't stared at the image long enough... – Mariano Suárez-Álvarez Oct 11 '15 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 '11 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 '11 at 7:49
• I hope you don't mind if I use both ideas in another answer. – Wok Sep 4 '11 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 '11 at 19:31

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 – leonbloy Sep 2 '11 at 23:24

I believe this illustration is due to Anders Kaseorg: 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 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 $1 + 2 + \ldots + n$.

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 '11 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 '11 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 '11 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 – mrs Jun 27 '12 at 11:11

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

• But you're relying on a fact that is (often) proved with induction? – user636532 Jun 26 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? – punctured dusk Jun 26 at 9:09

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 $n(n+1)$.

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 $(\lim_{x\to1}\frac{x^n(nx-n-1)+1}{x^2-2x+1})^2$. You can also prove it using telescopic series.

• why does the assumption hold? this is usually proved using induction... – akkkk Jun 27 '12 at 9:47
• what assumption?? – Aang Jun 27 '12 at 9:48
• I guess Auke means $\sum_{0}^{n} x^k=\frac{1-x^{n+1}}{1-x}$. – sdcvvc Jun 27 '12 at 10:12
• LHS is a geometric series. en.wikipedia.org/wiki/Geometric_progression – Aang Jun 27 '12 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 '12 at 12:40

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 '12 at 10:40
• @avatar: Dear Avatar; I will. Your proof is great and it looks analytically. Thanks. :-) – mrs Jun 27 '12 at 10:48
• sorry, if you felt that i was being rude. – Aang Jun 27 '12 at 10:53
• @avatar: Your proof lights mine. No Problem at all. WELCOME Avatar. :) :) – mrs Jun 27 '12 at 10:57

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

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

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

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

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

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

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

$=[1+1+2+2+3+3+4+4+\cdots+(n-1)+(n-1)+n][1-1+2-2+3-3+4-4+\cdots+(n-1)-(n-1)+n]=$

$=[2(1+2+3+4+\cdots+(n-1))+n]n=(2\frac{n(n-1)}{2}+n)n=(n(n-1)+n)n=n^3$

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 '11 at 13:26

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 $\sum k^2$. – punctured dusk Jul 6 at 7:47

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

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

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 $B_1'$.

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

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...

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

Notice, $$k(k-1)(k-2) = \frac {k!}{(k-3)!}$$ Now we have $$\sum_{k=0}^n \frac {k!}{(k-3)!} + \sum_{k=0}^n 3k^2 - \sum_{k=0}^n 2k$$ 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! $$\sum_{k=0}^n \frac {k!}{(k-3)!} = 3!\sum_{k=0}^n \frac {k!}{(k-3)!3!} = 3!\sum_{k=0}^n\binom{k}{3} = 3!\binom{n+1}{4}$$ 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 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 $$\boxed{\sum_{k=1}^n k^3=\left(\sum_{k=1}^n k\right)^2}$$

To the collection of the geometric proofs: