Nicomachus's Theorem

Theorem

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

In general:

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

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 $n \in \N^*$, 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 $T_{k+1}$:

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

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

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle T_{k+1}\)	\(=\)	\(\displaystyle T_k + k \left({2k}\right) + \left({k + 1}\right)^2 + \left({k + 1}\right) - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle k^3 + k \left({2k}\right) + \left({k + 1}\right)^2 + \left({k + 1}\right) - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle k^3 + 2k^2 + k^2 + 2k + 1 + k + 1 - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle k^3 + 3k^2 + 3k + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({k + 1}\right)^3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

Therefore:

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

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:

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

$\blacksquare$

Source of Name

This entry was named for Nicomachus of Gerasa.

Sources

• Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): Exercise $1.2.1: \ 8$