\[\begin{align*} E(X) &= \sum_{k\ge0} k(1-p)^kp\\ \text{let: } q &=1-p \\ &=p\sum_{k\ge0} kq^k \\ &= p\sum_{k\ge1} kq^k \text{ (since } k=0 \text{ term is } 0) \\ &= p\frac{q}{(1-q)^2} \\ &= \frac{pq}{p^2} \\ &= \frac{1-p}{p} \end{align*}\]
\[\begin{align*} E(X) &= \sum_{k\ge0} k(1-p)^{k-1}p\\ \text{let: } q &=1-p \\ &= p\sum_{k\ge0} kq^{k-1} \\ &= \frac{p}{q} \sum_{k\ge0} kq^k\\ &= \frac{p}{q} \sum_{k\ge1} kq^k \text{ (since } k=0 \text{ term is } 0) \\ &= \frac{p}{q}\frac{q}{(1-q)^2}\\ &= \frac{p}{p^2} \\ &= \frac{1}{p} \end{align*}\]
\[\begin{align*} E[X^2] &= \sum_{k\ge0} k^2(1-p)^kp \\ \text{let: } q &= 1-p \\ &= \sum_{k\ge0} k^2q^kp \\ &= p \sum_{k\ge0} k^2q^k \\ &= p \frac{q(1+q)}{(1-q)^3} \\ &= \frac{p(1-p)(1+(1-p))}{p^3} \\ &= \frac{(1-p)(2-p)}{p^2} \end{align*}\]
Then:
\[\begin{align*} Var(X) &= E[X^2] - (E[X])^2 \\ &= \frac{(1-p)(2-p)}{p^2} - \left(\frac{1-p}{p}\right)^2 \\ &= \frac{(1-p)(2-p) - (1-p)^2}{p^2} \\ &= \frac{(1-p)[2-p - 1+p]}{p^2} \\ &= \frac{1-p}{p^2} \end{align*}\]
\[\begin{align*} E[X^2] &= \sum_{k\ge0} k^2(1-p)^{k-1}p \\ \text{let: } q &= 1-p \\ &= \sum_{k\ge1} k^2q^{k-1}p \text{ (since } k=0 \text{ term is } 0) \\ &= \sum_{k\ge1} k(k+1)pq^{k-1} - \sum_{k\ge1}kpq^{k-1} \\ \text{and: } \sum_{k\ge1}kpq^{k-1} &= \frac{1}{p}\text{ (expected value of the shifted geometric distribution)} \\ &= \sum_{k\ge1} k(k+1)pq^{k-1} - \frac{1}{p} \\ &= \frac{2p}{(1-q)^3} - \frac{1}{p} \\ &= \frac{2p}{p^3} - \frac{1}{p} \\ &= \frac{2}{p^2} - \frac{1}{p} \\ \end{align*}\]
Then:
\[\begin{align*} Var(X) &= E[X^2] - (E[X])^2 \\ &= \frac{2}{p^2} - \frac{1}{p} - \left(\frac{1}{p}\right)^2 \\ &= \frac{2}{p^2} - \frac{1}{p} - \frac{1}{p^2} \\ &= \frac{1}{p^2} - \frac{1}{p} \\ &= \frac{1-p}{p^2} \end{align*}\]
\[\begin{align*} M_X(t) &= E[e^{tX}] \\ &=\sum_{k=0}^\infty e^{tk}(1-p)p^k \\ &= (1-p)\sum_{k=0}^\infty (pe^t)^k \\ &= (1-p)\frac{1}{1-pe^t}, \text{ for } pe^t<1 \\ &= \frac{1-p}{1-pe^t}, t<-\ln p \end{align*}\]
\[\begin{align*} M_X(t) &= E[e^{tX}] \\ &=\sum_{k=1}^\infty e^{tk}(1-p)^{k-1}p \\ &= p\sum_{k=1}^\infty e^{tk}q^{k-1} \\ &= \frac{p}{q}\sum_{k=1}^\infty e^{tk}q^k \\ &= \frac{p}{q}\sum_{k=1}^\infty (qe^t)^k \\ &= \frac{p}{q} \cdot \frac{qe^t}{1-e^t(1-p)} \\ &= \frac{pe^t}{1-e^t(1-p)}, t<-\ln(1-p) \end{align*}\]