Nicomachus's Theorem

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Theorem

\(\displaystyle 1^3\) \(=\) \(\displaystyle 1\)
\(\displaystyle 2^3\) \(=\) \(\displaystyle 3 + 5\)
\(\displaystyle 3^3\) \(=\) \(\displaystyle 7 + 9 + 11\)
\(\displaystyle 4^3\) \(=\) \(\displaystyle 13 + 15 + 17 + 19\)
\(\displaystyle \vdots\) \(\) \(\displaystyle \)


In general:

$\forall n \in \N_{>0}: n^3 = \left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 + n - 1}\right)$


In particular, the first term for $\left({n + 1}\right)^3$ is $2$ greater than the last term for $n^3$.


Proof 1

Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:

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


Basis for the Induction

$P(1)$ 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 $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.


So this is our induction hypothesis:

$k^3 = \left({k^2 - k + 1}\right) + \left({k^2 - k + 3}\right) + \ldots + \left({k^2 + k - 1}\right)$


Then we need to show:

$\left({k + 1}\right)^3 = \left({\left({k + 1}\right)^2 - \left({k + 1}\right) + 1}\right) + \left({\left({k + 1}\right)^2 - \left({k + 1}\right) + 3}\right) + \ldots + \left({\left({k + 1}\right)^2 + \left({k + 1}\right) - 1}\right)$


Induction Step

Let $T_k = \left({k^2 - k + 1}\right) + \left({k^2 - k + 3}\right) + \ldots + \left({k^2 + k - 1}\right)$.

We can express this as:

$T_k = \left({k^2 - k + 1}\right) + \left({k^2 - k + 3}\right) + \ldots + \left({k^2 - k + 2k - 1}\right)$

We see that there are $k$ terms in $T_k$.


Let us consider the general term $\left({\left({k + 1}\right)^2 - \left({k + 1}\right) + j}\right)$ in $T_{k+1}$:

\(\displaystyle \left({k + 1}\right)^2 - \left({k + 1}\right) + j\) \(=\) \(\displaystyle k^2 + 2 k + 1 - \left({k + 1}\right) + j\)
\(\displaystyle \) \(=\) \(\displaystyle k^2 - k + j + 2 k\)


So, in $T_{k+1}$, each of the terms is $2k$ larger than the corresponding term for $T_k$.

So:

\(\displaystyle T_{k+1}\) \(=\) \(\displaystyle T_k + k \left({2k}\right) + \left({k + 1}\right)^2 + \left({k + 1}\right) - 1\)
\(\displaystyle \) \(=\) \(\displaystyle k^3 + k \left({2k}\right) + \left({k + 1}\right)^2 + \left({k + 1}\right) - 1\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle k^3 + 2k^2 + k^2 + 2k + 1 + k + 1 - 1\)
\(\displaystyle \) \(=\) \(\displaystyle k^3 + 3k^2 + 3k + 1\)
\(\displaystyle \) \(=\) \(\displaystyle \left({k + 1}\right)^3\)


So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N_{>0}: n^3 = \left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 + n - 1}\right)$


Finally, note that the first term in the expansion for $\left({n + 1}\right)^3$ 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:

$\left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 + n - 1}\right)$

can be written:

$\left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 - n + 2 n - 1}\right)$

Writing this in sum notation:

\(\displaystyle \) \(\) \(\displaystyle \left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 - n + 2 n - 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n \left({n^2 - n + 2 k - 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle n \left({n^2 - n}\right) + \sum_{k \mathop = 1}^n \left({2 k - 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle n^3 - n^2 + n^2\) Odd Number Theorem
\(\displaystyle \) \(=\) \(\displaystyle n^3\)

$\blacksquare$


Source of Name

This entry was named for Nicomachus of Gerasa.