Nicomachus's Theorem

Theorem
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 $T_{k+1}$:

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

So:

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 $n^3$.

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: