Sum of Squared Deviations from Mean/Corollary 2

Theorem
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a set of real numbers.

Let $\overline x$ denote the arithmetic mean of $S$.

Then:
 * $\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n x_i^2 - \frac 1 n \paren {\sum_{i \mathop = 1}^n x_i}^2$

Proof
For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.

Then: