Jtdylktuy

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 $sumlimits_{k=1}^n k^3$.

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

Therefore, $sumlimits_{k=1}^n k^3=left(sumlimits_{k=1}^nkright)^2$

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.

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

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

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:

• 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
  

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 $n(n+1)$.

Here's a direct algebraic proof. $$sum_{k=1}^n(k^3-k^2)=2sum_{k=1}^nkcdotfrac{k(k-1)}2=2sum_{k=1}^nksum_{l=1}^{k-1}l=2sum_{1leqslant l<kleqslant n}kl=left(sum_{k=1}^nkright)^2-sum_{k=1}^nk^2$$

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 $lim_{xto1}$ 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 $(lim_{xto1}frac{x^n(nx-n-1)+1}{x^2-2x+1})^2$. You can also prove it using telescopic series.

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

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

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

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

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

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

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

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

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.

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)

The square in the identity is the area of the triangle below, while the cubes are the area's 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:
$(1 + 2 + cdots + n)^2 = sum_{k=1}^n k^3$ , $blacksquare$

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
$$begin{align}
R_6 &= 6+12+18+24+30+36+42+48+54\
&= 6(1+2+3+4+5+6+7+8+9)\
&= 6sum_{i=1}^9 i
end{align}$$
And the sum of all the entries becomes
$$sum_{i=1}^n R_i = sum_{i=1}^n bigg(isum_{j=1}^n jbigg)=bigg(sum_{j=1}^n jbigg)bigg(sum_{i=1}^n ibigg)=bigg(sum_{i=1}^n ibigg)^2$$
Thus we have
$$sum_{i=1}^n i^3 = sum_{i=1}^n L_i = sum_{i=1}^n R_i=bigg(sum_{i=1}^n ibigg)^2$$

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 ${(i,j) in mathbb{Z}timesmathbb{Z}: i + j leq n, i > 0, j > 0}$.

Then $left(sum_{k=1}^n kright)^2$ 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 $B_1'$.

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...

For every $kinmathbb{N}$
$$(k+1)^4=k^4+4k^3++6k^2+4k+1$$
therefore
$$sum_{k=1}^n(k+1)^4=sum_{k=1}^nk^4+4sum_{k=1}^nk^3+6sum_{k=1}^nk^2+4sum_{k=1}^nk+sum_{k=1}^n1$$
which is equivalent to
$$sum_{k=1}^nk^4+(n+1)^4-1=sum_{k=1}^nk^4+4sum_{k=1}^nk^3+6sum_{k=1}^nk^2+2n(n+1)+n$$
After simplifications we obtain
$$4sum_{k=1}^nk^3=(n+1)^4-1-2n(n+1)-n-6sum_{k=1}^nk^2=n^4+4n^3+4n^2+n-6sum_{k=1}^nk^2$$
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
$$4sum_{k=1}^nk^3=n^4+4n^3+4n^2+n-6sum_{k=1}^nk^2\=n^4+4n^3+4n^2+n-n(n+1)(2n+1)\=n^4+2n^3+n^2=n^2(n+1)^2$$
Finally
$$sum_{k=1}^nk^3=frac{n^2(n+1)^2}{4}=Big(frac{n(n+1)}{2}Big)^2=Big(sum_{k=1}^nkBig)^2$$

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 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!
$$sum_{k=0}^n frac {k!}{(k-3)!} = 3!sum_{k=0}^n frac {k!}{(k-3)!3!} = 3!sum_{k=0}^nbinom{k}{3} = 3!binom{n+1}{4}$$
We have now obtained
$$sum_{k=0}^n k^3 = 3!binom{n+1}{4} + 3sum_{k=0}^n k^2 - 2sum_{k=0}^n k$$
Focusing solely on the right-hand side we have
$$6biggl(frac {(n+1)!}{(n-3)!4!}biggr) + 3sum_{k=0}^n k^2 - 2sum_{k=0}^n k$$
Assuming we already know the sum of the sequence of integers and squared integers (the 2 sums we have left) we have
$$ frac {(n+1)(n)(n-1)(n-2)}{4} + 3frac{n(n+1)(2n+1)}{6} - 2frac {n(n+1)}{2}$$
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:
$$ frac {n^4+2n^3+n^2}{4} = biggl(sum_{k=0}^n kbiggr)^2$$