# Alternating Sum and Difference of Sequence of Squares

## Theorem

 $\ds \sum_{j \mathop = 1}^{2 n} \paren {-1}^{j - 1} j^2$ $=$ $\ds 1^2 - 2^2 + 3^2 - \cdots + \paren {2 n - 1} - \paren {2 n}^2$ $\ds$ $=$ $\ds -n \paren {2 n + 1}$

## 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:

 $\ds \sum_{j \mathop = 1}^2 \paren {-1}^{j - 1} j^2$ $=$ $\ds 1^2 - 2^2$ $\ds$ $=$ $\ds -3$ $\ds$ $=$ $\ds -1 \paren {2 \times 1 + 1}$

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:

 $\ds \sum_{j \mathop = 1}^{2 \paren {k + 1} } \paren {-1}^{j - 1} j^2$ $=$ $\ds \sum_{j \mathop = 1}^{2 k} \paren {-1}^{j - 1} j^2 + \paren {2 \paren {k + 1} - 1}^2 - \paren {2 \paren {k + 1} }^2$ $\ds$ $=$ $\ds -k \paren {2 k + 1} + \paren {2 k + 1}^2 - \paren {2 k + 2}^2$ Induction hypothesis and some simplification $\ds$ $=$ $\ds -2 k^2 - k + 4 k^2 + 4 k + 1 - 4 k^2 - 8 k - 4$ multiplying out $\ds$ $=$ $\ds -2 k^2 - 5 k - 3$ $\ds$ $=$ $\ds -\paren {k + 1} \paren {2 k + 3}$ $\ds$ $=$ $\ds -\paren {k + 1} \paren {2 \paren {k + 1} + 1}$

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

Therefore:

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

$\blacksquare$