Variance of Shifted Geometric Distribution/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:
 * $\displaystyle \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \, \map \Pr {X = x}$

To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$.

Thus:

Then: