$\textbf{Nicomachus}$ discovered “Nicomachus Theorem” interesting number patterns involving cubes and sums of odd numbers. Nicomachus was born in Roman Syria (now Jerash, Jordan) around 100 AD. He wrote in Greek as a Pythagorean.
$\textbf{Nicomachus Theorem: Cubes and Sums of Odd numbers}$
\begin{eqnarray*}
1 &=& 1^3 \\
3 + 5 &=& 8 = 2^3 \\
7 + 9 + 11 &=& 27 = 3^3 \\
13 + 15 + 17 + 19 &=& 64 = 4^3 \\
21 + 23 + 25 + 27 + 29 &=& 125 = 5^3 \\
&\vdots&
\end{eqnarray*}
In general, $(n^2-n+1)+(n^2-n+3)+\cdots+(n^2-n+2n-1)=n^3$.
In particular, the first term for $(n+1)^3$ is $2$ greater than the last term for $n^3$.
- $3$ is $2$ greater than $1$
- $7$ is $2$ greater than $5$
- $13$ is $2$ greater than $11$
- $21$ is $2$ greater than $19$
The $n^{\text{th}}$ cubic number $n^3$ is a sum of $n$ consecutive odd numbers.
The identity is;
$$\sum_{k=1}^n \big[n(n-1)+2k-1\big] = n^3$$
$\textit{Proof}$
\( \begin{aligned} \displaystyle
\sum_{k=1}^n \big[n(n-1)+2k-1\big] &= \sum_{k=1}^n (n^2-n + 2k-1) \\
&= \sum_{k=1}^n (n^2-n) + \sum_{k=1}^n {2k}-\sum_{k=1}^n {1} \\
&= n(n^2-n) + 2(1+2+3+\cdots+n)-(1+1+1+\cdots+1) \\
&= n(n^2-n) + 2 \times \dfrac{n}{2}(1+n)-n \\
&= n^3-n^2 + n + n^2-n\\
&= n^3
\end{aligned} \)
$\textbf{Nicomachus Theorem: How Squares and Cubes Meet}$
\begin{eqnarray*}
1^3 &=& 1^2 \\
1^3 + 2^3 &=& (1 + 2)^2 \\
1^3 + 2^3 + 3^3 &=& (1 + 2 + 3)^2 \\
1^3 + 2^3 + 3^3 + 4^3 &=& (1 + 2 + 3 + 4)^2 \\
\vdots \\
1^3 + 2^3 + 3^3 + \cdots + n^3 &=& (1 + 2 + 3 + \cdots + n)^2
\end{eqnarray*}
$\textit{Proof by Mathematical Induction}$
Step 1: Show it is true for $n=1$.
$\text{LHS} = 1^3 = 1 $
$\text{RHS} = 1^2 = 1$
$\therefore$ It is true for $n=1$.
Step 2: Assume it is true for $n=k$.
That is, $1^3 + 2^3 + \cdots + k^3 = (1+2+\cdots+k)^2$.
Step 3: Show it is true for $n=k+1$.
That is, $1^3 + 2^3 + \cdots + k^3 + (k+1)^3 = (1+2+\cdots+k+k+1)^2$.
\( \begin{aligned} \displaystyle
\text{LHS} &= 1^3 + 2^3 + \cdots + k^3 + (k+1)^3 \\
&= (1+2+\cdots+k)^2 + (k+1)^3 \\
&= \Big[\dfrac{k}{2}(1+k)\Big]^2 + (k+1)^3 \\
&= \dfrac{k^2}{4}(1+k)^2 + (k+1)^3 \\
&= (1+k)^2\Big[\dfrac{k^2}{4}+(k+1)\Big] \\
&= (1+k)^2 \times \dfrac{k^2+4k+4}{4} \\
&= \dfrac{(1+k)^2}{4}(k+2)^2 \\
&= \dfrac{(1+k)^2}{4}(1+k+1)^2 \\
&= \Big[\dfrac{1+k}{2}(1+k+1)\Big]^2 \\
&= (1+2+\cdots+k+k+1)^2 \\
&= \text{RHS}
\end{aligned} \)
$\textit{Proof by Geometry}$
The picture shows;
- the area of the red square is $1=1^3$.
- the area of the yellow square is $3^2-1=8=2^3$.
- the area of the blue square is $6^2-3^2=27=3^3$.
- the area of the green square is $10^2-6^2=647=4^3$.
So the total area is the sum of consecutive cubes, which is $1^3 + 2^3 + 3^3 + 4^3$.
Reading along the top edge, we find $1+2+3+4$, the sum of consecutive numbers. But the area of a square is the square of the length of its side, which is $(1+2+3+4)^2$.
\( \begin{align}
1^3 + 2^3 + 3^3 + 4^3 &= (1+2+3+4)^2 \\
\therefore 1^3 + 2^3 + 3^3 + \cdots + n^3 &= (1+2+3+ \cdots + n)^2 \\
\end{align} \)
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