Proving the identity $\sum\limits_{k=1}^n {k^3} = \left(\sum\limits_{k=1}^n k\right)^2$ without induction

Question

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.

Thanks!

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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 $1 + 2 + \ldots + n$.

Answer 6

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

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

Answer 7

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.

Answer 8

I found this but sadly do not understand it, can someone explain the method?

Edit:

This has the same proof but much easier to understand. It is also found in the link by QY

Answer 9

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

Answer 10

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.

Answer 11

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.

Answer 12

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 $n(n+1)$.

Answer 13

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

Answer 14

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)

Answer 15

I believe this illustration is due to Anders Kaseorg:

Answer 16

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.

Answer 17

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

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

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

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

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

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

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

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

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