Using expectation proof of mean and variance of geometric distribution?

The mean of a geometric distribution, which models the number of trials until the first success, can be derived by considering the expected value (E[X]) as (E[X] = \sum_{k=1}^{\infty} k \cdot P(X = k) = \sum_{k=1}^{\infty} k \cdot (1-p)^{k-1} p). By using the formula for the sum of a geometric series and differentiating, we find that the mean is ( \frac{1}{p} ). For the variance, we first calculate (E[X^2]) and then use the formula (Var(X) = E[X^2] - (E[X])^2), resulting in (Var(X) = \frac{1-p}{p^2}).