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

Question

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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?

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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

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

Answer 5

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:

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