Variance of Shifted Geometric Distribution/Proof 1

Theorem
Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.

Then the variance of $X$ is given by:
 * $\displaystyle \operatorname{var} \left({X}\right) = \frac {1-p} {p^2}$

Proof
From the definition of Variance as Expectation of Square minus Square of Expectation:
 * $\operatorname{var} \left({X}\right) = E \left({X^2}\right) - \left({E \left({X}\right)}\right)^2$

From Expectation of Function of Discrete Random Variable:
 * $\displaystyle E \left({X^2}\right) = \sum_{x \mathop \in \operatorname{Im} \left({X}\right)} x^2 \Pr \left({X = x}\right)$

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

Thus:

Then: