4.1.13 problem 14

$\text{E}\left(X\right)=c{\sum }_{i=1}^{\infty }{p}^k=c(\frac{1}{1-p}-1)=-\frac{1}{\log (1-p)}\frac{p}{1-p}$

$\text{E}\left(X\right)=\text{E}\left(X^2\right)-{\left(\text{E}\right(X\left)\right)}^2=c{\sum }_{i=1}^{\infty }i{p}^i-{(-\frac{1}{\log (1-p)}\frac{p}{1-p})}^2=-\frac{1}{\log (1-p)}\frac{p}{{(p-1)}^2}-{(-\frac{1}{\log (1-p)}\frac{p}{1-p})}^2$