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:
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, $\sum\limits_{k=1}^n k^3=\left(\sum\limits_{k=1}^nk\right)^2$
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.
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. $$\sum_{k=1}^n(k^3-k^2)=2\sum_{k=1}^nk\cdot\frac{k(k-1)}2=2\sum_{k=1}^nk\sum_{l=1}^{k-1}l=2\sum_{1\leqslant l<k\leqslant n}kl=\left(\sum_{k=1}^nk\right)^2-\sum_{k=1}^nk^2$$
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.
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: $$S_k^2 - S_{k-1}^2 = k \left(k+2 S_{k-1}\right) = k\left(k+k\left(k-1\right)\right)=k^3$$
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:
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 $$\boxed{\sum_{k=1}^n k^3=\left(\sum_{k=1}^n k\right)^2}$$
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.
$f(n)=1^3+2^3+3^3+\cdots+n^3$
$f(n-1)=1^3+2^3+3^3+\cdots+(n-1)^3$
$f(n)-f(n-1)=n^3$
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 $a^2-b^2=(a+b)(a-b)$
$$\begin{align}g(n)-g(n-1)&=\\ &=[(1+\dots+n)+(1+\dots +(n-1))][(1+\dots+ n)-(1+\dots+(n-1)]\\ &=[2(1+2+3+4+\cdots+(n-1))+n]n\\ &=\left(2\frac{n(n-1)}{2}+n\right)n\\ &==(n(n-1)+n)n\\ &=n^3 \end{align}$$
$f(n)-f(n-1)=g(n)-g(n-1)=n^3$ for all $n$ and if $f(0)=g(0)$ then
$f(n)$ and $g(n)$ are equal and we can write $$1^3+2^3+3^3+\cdots+n^3=(1+2+3+4+\cdots+n)^2$$
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 $$B=\sum_1^n k=\frac{1}{2}n^2+\frac{1}{2}n$$ $A-B^2=0$. :)
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 $$\sum_{i=1}^n i^3 = \sum_{i=1}^n L_i = \sum_{i=1}^n R_i=\bigg(\sum_{i=1}^n i\bigg)^2$$
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.
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: $(1 + 2 + \cdots + n)^2 = \sum_{k=1}^n k^3$ , $\blacksquare$
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:
$$(\sum k)^2=\sum j^2 + 2 \sum_{i<j} ij =\sum_j (j^2+2 \sum_{i<j} ij)=\sum_j j(j+j(j-1))=\sum_j j(j^2)=\sum_j j^3$$
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 $j$.
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 $S_3(x) = S_1(x)^2$.
