# Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”

This site states:

Example $$\boldsymbol 3$$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$ “The sum of $$n$$ consecutive cubes is equal to the square of the sum of the first $$n$$ numbers.”

In other words, according to Example $$1$$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$

Should: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$ not be: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$ as everything in the left-hand side is cubed?

• Why should it be cubed just because every integer to the left is cubed? That is exactly the point of this example. Do you know that $\sum\limits_{k=1}^n k = 1+2+3+\dots + n = \frac{n(n+1)}{2}$? Jun 17, 2015 at 12:34
• The easiest way to check this is to try the formula out for $n=1$ - do you want $1^3=1$ as the formula you were given computes, or $1^3=\frac 98$ as you get applying your suggested alternative? Jun 17, 2015 at 12:36
• $(a+b)^3 = a^3 + 3ab^2 + 3a^2 b + b^3 \neq a^3 + b^3$ Jun 17, 2015 at 13:45

No! Generally speaking, one shows by induction that $\,1^r+2^r+\dots+n^r\,$ has a closed form which is a polynomial in $n$ of degree $\color{red}{r+1}$.

Examples:

1. $1 +2 +\dots+n =\dfrac{n(n+1)}2$.
2. $1^2+2^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}6$
3. $1^0+2^0+\dots+n^0=\underbrace{1+1++\dots+1}_{n \ \text{times}}=n$

and the formula you posted about. What you propose hasn't the required degree, so it can't be true.

Argument:

Every $$k^3$$ is the sum of $$k$$ consecutive odd numbers.
E.g.: $$8 = 3+5$$, $$\quad 27 = 7+9+11$$, $$\quad 64 = 13+15+17+19$$.

The sum of $$n$$ consecutive $$k^3$$ numbers, starting from $$k = 1$$, is the sum of
$$n(n+1)/2$$ consecutive odd numbers.

But the sum of a number of consecutive odd numbers is the square of that number.

Therefore, the sum of $$n$$ consecutive $$k^3$$ starting with $$1$$ is $$[n(n+1)/2]^2$$.

Elaboration:

It is well known that $$k^2$$ is the sum of $$k$$ consecutive odd numbers starting from $$1$$. But $$k^3$$, $$k^4$$, and $$k^p$$ generally, are also the sum of $$k$$ consecutive odd numbers. This is because $$k$$ numbers whose arithmetic mean is $$k$$ sum to $$k^2$$; $$k$$ numbers whose mean is $$k^2$$ sum to $$k^3$$, et cetera. What is unique about the sum of cubes is that it is a sum of consecutive odd numbers with no gaps and no repetitions. E.g., the sum of the first five cubes is:

$$\underbrace{1^3}_{1} +\underbrace{2^3}_{3+5} +\underbrace{3^3}_{7+9+11} +\underbrace{4^3}_{13+15+17+19} +\underbrace{5^3}_{21+23+25+27+29}.$$

Therefore the sum is a square. But what square? In summing the first five cubes we have summed the first $$1+2+3+4+5 = 15$$ odd numbers, giving us $$15^2$$. And $$15$$ is the fifth triangle number. Hence generally, the sum of the first $$n$$ cubes is the square of the $$n$$th triangle number, or $$[n(n+1)/2]^2$$, i.e. the square of the sum of the first $$n$$ integers.

With sums of squares, there is repetition because all the sums of odd numbers start from $$1$$:

$$1^2 +2^2 +3^2 +4^2 = 1 +(1+3) +(1+3+5) +(1+3+5+7).$$

With fourth and higher powers there is no repetition, but there are gaps between the sets of odd numbers being summed:

$$1^4 +2^4 +3^4 +4^4 = 1 +(7+9) +(25+27+29) +(61+63+65+67).$$

Only sums of cubes are sums of consecutive odd numbers starting from $$1$$, and therefore have this special connection with squares.

Consider the infinite "multiplication table" array $$M$$ defined by $$M_{i,j}=ij$$

$$M = \left[\begin{array}{r} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 6 & 8 & 10 \\ 3 & 6 & 9 & 12 & 15 & \cdots\\ 4 & 8 & 12 & 16 & 20 \\ 5 & 10 & 15 & 20 & 25 \\ && \vdots &&& \ddots \end{array}\right]$$

Let $$L(n) = \{M_{i,j} : 1 \le i \le n \wedge j=n \} \cup \{M_{i,j} : i=n \wedge 1 \le j \le n \}$$

be sum of the the _| shaped region consisting of all elements of $$M$$ whose first or second index is $$n$$ and whose other index is less than or equal to $$n$$. For example, the elements of $$L(4)$$ are shown below.

$$M = \left[\begin{array}{r} 1 & 2 & 3 & \color{red}4 & 5 \\ 2 & 4 & 6 & \color{red}8 & 10 \\ 3 & 6 & 9 & \color{red}{12} & 15 & \cdots\\ \color{red}4 & \color{red}8 & \color{red}{12} & \color{red}{16} & 20 \\ 5 & 10 & 15 & 20 & 25 \\ && \vdots &&& \ddots \end{array}\right]$$

It turns out that $$L(1)=1,\; L(2)=8,\; L(3)=27,\; L(4)=64,$$ and $$L(5)=125$$.

We show that $$L(n) = n^3$$.

\begin{align} L(n) &= \sum_{i=1}^n M_{i,n} + \sum_{j=1}^n M_{n,j} - M_{n,n} \\ &= n\sum_{i=1}^n i + n\sum_{j=1}^n j - n^2 \\ &= n^2(n+1) - n^2 \\ &= n^3 \end{align}

Let $$S(n)$$ be the sum of the $$n^2$$ elements in the $$n \times n$$ subarray of elements in the upper-left corner of $$M$$.

For example, $$S(3) = (1+2+3) + (2 + 4 + 6) + (3 + 6 + 9) = 36$$

We will compute $$S(n)$$ two ways.

\begin{align} \sum_{i=1}^n \sum_{j=1}^n M_{i,j} &= \sum_{i=1}^n \sum_{j=1}^n ij \\ &= \left(\sum_{i=1}^n i \right) \left(\sum_{j=1}^n j \right) \\ &=\left[\frac{n(n+1)}{2}\right]^2 \end{align}

Also, \begin{align} S(n) &= L(1) + L(2) + \cdots + L(n) \\ S(n) &= \sum_{i=1}^n i^3 \end{align}

So $$\displaystyle \sum_{i=1}^n i^3 = \left[\frac{n(n+1)}{2}\right]^2$$

• This is a clever way to simply. Also good use of matrices. Upvoted.
– xax
Jan 6, 2022 at 11:22

$$1^3 +2^3 +\ldots +n^3 =\left(\frac{n(n+1)}{2}\right)^2.$$

We will prove by induction on $$n$$, that $$\sum_{k=1}^n k^3 = 1^3 +2^3 +\ldots +n^3 =\frac{n^2 (n+1)^2}{4}.$$

For $$n=1$$, we have $$1^3 = \dfrac{1^2 2^2}{4} = 1$$.

We shall prove that $$\sum_{k=1}^{n+1} k^3 = 1^3 +2^3 +\ldots +n^3 +(n+1)^3 =\frac{(n+1)^2 (n+2)^2}{4},$$ by assuming that $$\sum_{k=1}^n k^3 = 1^3 +2^3 +\ldots +n^3 =\frac{n^2 (n+1)^2}{4}.$$ From the induction supposition, we have to prove that $$\frac{n^2 (n+1)^2}{4} +(n+1)^3 = \frac{(n+1)^2 (n+2)^2}{4},$$ or $$(n+1)^3 = \frac{(n+1)^2 (n+2)^2}{4} -\frac{n^2 (n+1)^2}{4} = \frac{(n+1)^2}{4} ((n+2)^2 -n^2) = (n+1)^3.$$

Well you have:

$$1^3+\ldots+n^3=\frac{n^2(n+1)^2}{4}$$

You can prove it easily by induction.

Edit : One can prove the formula arranging squares thusly :

For any $$0\leq k\leq n$$, you can arrange $$4\times k$$ squares of $$k$$ by $$k$$ so that they form a square of $$n\times (n+1)$$.

Now your proposition could not work without even knowing the formula for the following reason:

$$\frac{n^3+(n+1)^3}{2}\text{ is equivalent to } n^3$$

Whereas I claim that:

$$\int_0^{n}x^3dx \leq \sum_{k=1}^nk^3\leq \int_0^{n+1}x^3dx$$

From which it follows that:

$$\sum_{k=1}^nk^3\text{ is equivalent to } \frac{n^4}{4}$$

• Your use of "equivalent" is a little obscure. Probably what you mean is that $f(n)$ and $g(n)$ are "equivalent" if and only if $\lim_{n\to +\infty} f(n)/g(n) = 1$. Jun 17, 2015 at 20:26
• @hardmath, you are right, this is what I meant, thanks for the clarification. Jun 18, 2015 at 6:49
• @ClémentGuérin any method other than induction?
– Ace
Aug 12 at 8:53
• @Ace, there are other methods. For instance one can use power series or some geometric argument. See my edit. Sep 20 at 4:58