Nicomachus's Theorem/Proof 1

Proof
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 $T_{k+1}$:

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

So:

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 $n^3$.