An interesting sum

Problem:

$1^2 - 2^2 + 3^2 - 4^2 + \cdots - (2n)^2$

Solution:

We will assume the formula:

$1 + 2 + \cdots + n = \frac{n(n+1)}{2}$

$1^2 + 2^2 + \cdots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$

We split the sum into two halves:

$1^2 + 3^2 + 5^2 + \cdots + (2n - 1)^2$ can be think of as

$(2 \times 0 + 1)^2 + (2 \times 1 + 1)^2 + \cdots + (2 \times (n - 1) + 1)^2$

Expanding them we get

$(4 \times 0^2 + 4 \times 0 + 1) + (4 \times 1^2 + 4 \times 1 + 1) + \cdots + (4(n-1)^2 + 4(n-1) + 1)$.

Grouping terms, factoring and applying the formula, we get

$4\frac{(n - 1)(n)(2n - 1)}{6} + 4\frac{(n-1)(n)}{2} + n$

Simplifying we get

$\frac{n(2n - 1)(2n + 1)}{3}$

The even number squared sum is easier, we can think of them as just a scaled version of the square sum

$2^2 + 4^2 + \cdots + (2n)^2$

$4(1^2 + 2^2 + \cdots + n^2)$

The answer is simply $\frac{2n(n + 1)(2n + 1)}{3}$

Therefore, the final answer is simply the difference of the two sums, and it is

$-n(2n + 1)$