Triangular Number as Alternating Sum and Difference of Squares

Theorem
Thus the $n$th triangular number can be expressed as the alternating sum and difference of squares:

So:

and so on.

Proof
The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \frac {n \left({n + 1}\right)} 2 = \sum_{j \mathop = 0}^{n - 1} \left({-1}\right)^j \left({n - j}\right)^2$

Basis for the Induction
$P \left({1}\right)$ is the case:

which is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown 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 the induction hypothesis:
 * $\displaystyle \frac {k \left({k + 1}\right)} 2 = \sum_{j \mathop = 0}^{k - 1} \left({-1}\right)^j \left({n - j}\right)^2$

from which it is to be shown that:
 * $\displaystyle \frac {\left({k + 1}\right) \left({k + 2}\right)} 2 = \sum_{j \mathop = 0}^k \left({-1}\right)^j \left({k - j + 1}\right)^2$

Induction Step
This is the induction step:

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