Alternating Sum and Difference of Sequence of Squares

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $\ds \forall n \ge 1: \sum_{j \mathop = 1}^{2 n} \paren {-1}^{j - 1} j^2 = -n \paren {2 n + 1}$

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to be true.

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:
 * $\ds \sum_{j \mathop = 1}^{2 k} \paren {-1}^{j - 1} j^2 = -k \paren {2 k + 1}$

Then we need to show:
 * $\ds \sum_{j \mathop = 1}^{2 \paren {k + 1} } \paren {-1}^{j - 1} j^2 = -\paren {k + 1} \paren {2 \paren {k + 1} + 1}$

Induction Step
This is our induction step:

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

Therefore:
 * $\ds \forall n \ge 1: \sum_{j \mathop = 1}^{2 n} \paren {-1}^{j - 1} j^2 = -n \paren {2 n + 1}$