The sum of squared deviations formula is:
\[ \sum (x_i - \overline{x})^2 \]
Expanding:
\[ \sum (x_i^2 - 2 x_i \overline{x} + \overline{x}^2) \]
Using the distributive property:
\[ \begin{gather} \sum (x_i^2) - \left( \sum 2 x_i \overline{x} \right) + \sum (\overline{x}^2) \\ \sum (x_i^2) - 2\overline{x} \left( \sum x_i \right) + \sum (\overline{x}^2) \end{gather}\]
Assume there are \(n\) terms:
\[ \sum (x_i^2) - 2\overline{x} \left( \sum x_i \right) + n(\overline{x}^2) \]
Since \( (\sum x_i) / n = \overline{x}\):
\[ \begin{gather} \sum (x_i^2) - 2\overline{x} \left( n \overline{x} \right) + n(\overline{x}^2) \\ \sum (x_i^2)- n (\overline{x}^2 ) \\ \sum (x_i^2)- n \left(\frac{\sum x_i}{n}\right)^2 \\ \sum (x_i^2)- \frac{(\sum x_i) ^2}{n} \end{gather}\]