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.