Intereting Posts

Why is the axiom of choice separated from the other axioms?
How does one show that $\int_{0}^{\pi/4}\sin(2x)\ln{(\ln^2{\cot{x}})}\mathrm dx=\ln{\pi\over4}-\gamma?$
Sum of the sum of the sum of the first $n$ natural numbers
Does convergence in $L^{p}$ implies convergence almost everywhere?
Prove that $\{\frac 1 n \mid n \in \mathbb N\} \cup \{0\}$ is closed in $\mathbb R$
Closed Convex sets of $\mathbb{R^2}$
basic differential forms
negative number divided by positive number, what would be remainder?
Why is the Kleene star of a null set is an empty string?
What all maths do I need to know to become good at machine learning.
Archimedean places of a number field
How to find $\int\frac{\sin x}{x}dx$
How to express $z^8 − 1$ as the product of two linear factors and three quadratic factors
Between bayesian and measure theoretic approaches
Is Koch snowflake a continuous curve?

I recently proved that

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

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.

- How find the sum of the last two digits of $(x^{2})^{2013} + \frac{1}{(x^{2})^{2013}}$ for $x + \frac{1}{x} = 3$?
- Proof: Series converges $\implies $ the limit of the sequence is zero
- Arithmetic progression
- Find a closed form for this infinite sum: $ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+ \dots$
- A sequence with infinitely many radicals: $a_{n}=\sqrt{1+\sqrt{a+\sqrt{a^2+\cdots+\sqrt{a^n}}}}$
- How can $\lim_{n\to \infty} (3^n + 4^n)^{1/n} = 4$?
- Prove $S \doteq \sum_{n=1}^{\infty} p_n < \infty \to \prod_{n=1}^{\infty} (1-p_n) > 0$ assuming $0 \leq p_n < 1$.
- Minimum value of given expression
- Simple algebra question - proving $a^2+b^2 \geqslant 2ab$
- Conjecture: the sequence of sums of all consecutive primes contains an infinite number of primes

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.

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:

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

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

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

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.

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

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

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

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

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

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

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

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

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)

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 $B_1’$.

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)\\

&= 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

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

- List of powers of Natural Numbers
- Prime Harmonic Series $\sum\limits_{p\in\mathbb P}\frac1p$
- Prove $\det(kA)=k^n\det A$
- If $(x_n) \to x$ then $(\sqrt{x_1x_2\cdots x_n}) \to x$
- Summing (0,1) uniform random variables up to 1
- Is the axiom of choice needed to show that $a^2=a$?
- Cylinder object in the model category of chain complexes
- Calculating the variance of the ratio of random variables
- Is there a function with infinite integral on every interval?
- How to find the parametric equation of a cycloid?
- Prove: If a sequence converges, then every subsequence converges to the same limit.
- How to determine in polynomial time if a number is a product of two consecutive primes?
- Transitive closure
- Mean Value property for harmonic functions on regions other than balls/spheres
- Prove that largest root of $Q_k(x)$ is greater than that of $Q_j(x)$ for $k>j$.