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 $\map P n$ be the proposition:
 * $\ds \frac {n \paren {n + 1} } 2 = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j \paren {n - j}^2$

Basis for the Induction
$\map P 1$ 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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \frac {k \paren {k + 1} } 2 = \sum_{j \mathop = 0}^{k - 1} \paren {-1}^j \paren {n - j}^2$

from which it is to be shown that:
 * $\ds \frac {\paren {k + 1} \paren {k + 2} } 2 = \sum_{j \mathop = 0}^k \paren {-1}^j \paren {k - j + 1}^2$

Induction Step
This is the induction step:

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