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?
 A: 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:


*

*$1 +2 +\dots+n =\dfrac{n(n+1)}2$.

*$1^2+2^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}6$

*$1^0+2^0+\dots+n^0=\underbrace{1+1++\dots+1}_{n \ \text{times}}=n$


and the formula you posted about. What you propose hasn't the required degree, so it can't be true.
A: 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.
A: 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 $\displaystyle \sum_{i=1}^n i^3 = \left[\frac{n(n+1)}{2}\right]^2$
A: $$1^3 +2^3 +\ldots +n^3 =\left(\frac{n(n+1)}{2}\right)^2.$$
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
$$(n+1)^3 = \frac{(n+1)^2 (n+2)^2}{4} -\frac{n^2 (n+1)^2}{4} = \frac{(n+1)^2}{4} ((n+2)^2 -n^2) = (n+1)^3.$$
A: Well you have:
$$1^3+\ldots+n^3=\frac{n^2(n+1)^2}{4} $$
You can prove it easily by induction. 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:
$$\sum_{k=1}^nk^3\text{ is equivalent to } \frac{n^4}{4}$$
