Nicomachus's Theorem
Theorem
\(\ds 1^3\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds 2^3\) | \(=\) | \(\ds 3 + 5\) | ||||||||||||
\(\ds 3^3\) | \(=\) | \(\ds 7 + 9 + 11\) | ||||||||||||
\(\ds 4^3\) | \(=\) | \(\ds 13 + 15 + 17 + 19\) | ||||||||||||
\(\ds \vdots\) | \(\) | \(\ds \) |
In general:
- $\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
In particular, the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.
Proof 1
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
- $n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
Basis for the Induction
$\map P 1$ is true, as this just says $1^3 = 1$.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $k^3 = \paren {k^2 - k + 1} + \paren {k^2 - k + 3} + \dotsb + \paren {k^2 + k - 1}$
Then we need to show:
- $\paren {k + 1}^3 = \paren {\paren {k + 1}^2 - \paren {k + 1} + 1} + \paren {\paren {k + 1}^2 - \paren {k + 1} + 3} + \dotsb + \paren {\paren {k + 1}^2 + \paren {k + 1} - 1}$
Induction Step
Let $T_k = \paren {k^2 - k + 1} + \paren {k^2 - k + 3} + \dotsb + \paren {k^2 + k - 1}$.
We can express this as:
- $T_k = \paren {k^2 - k + 1} + \paren {k^2 - k + 3} + \dotsb + \paren {k^2 - k + 2k - 1}$
We see that there are $k$ terms in $T_k$.
Let us consider the general term $\paren {\paren {k + 1}^2 - \paren {k + 1} + j}$ in $T_{k+1}$:
\(\ds \paren {k + 1}^2 - \paren {k + 1} + j\) | \(=\) | \(\ds k^2 + 2 k + 1 - \paren {k + 1} + j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds k^2 - k + j + 2 k\) |
So, in $T_{k + 1}$, each of the terms is $2 k$ larger than the corresponding term for $T_k$.
So:
\(\ds T_{k + 1}\) | \(=\) | \(\ds T_k + k \paren {2 k} + \paren {k + 1}^2 + \paren {k + 1} - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds k^3 + k \paren {2 k} + \paren {k + 1}^2 + \paren {k + 1} - 1\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds k^3 + 2 k^2 + k^2 + 2 k + 1 + k + 1 - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds k^3 + 3 k^2 + 3 k + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1}^3\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
Finally, note that the first term in the expansion for $\paren {n + 1}^3$ is $n^2 - n + 1 + 2 n = n^2 + n + 1$.
This is indeed two more than the last term in the expansion for $n^3$ .
$\blacksquare$
Proof 2
From the definition:
- $\paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$
can be written:
- $\paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 - n + 2 n - 1}$
Writing this in sum notation:
\(\ds \) | \(\) | \(\ds \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 - n + 2 n - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {n^2 - n + 2 k - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \paren {n^2 - n} + \sum_{k \mathop = 1}^n \paren {2 k - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n^3 - n^2 + n^2\) | Odd Number Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds n^3\) |
$\blacksquare$
Source of Name
This entry was named for Nicomachus of Gerasa.