Variance of Geometric Distribution/Formulation 2/Proof 1

Proof
From the definition of Variance as Expectation of Square minus Square of Expectation:
 * $\var X = \expect {X^2} - \paren {\expect X}^2$

From Expectation of Function of Discrete Random Variable:
 * $\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$


 * Let $q = 1 - p$

Thus:

Then: