It is Question #35 from this https://math.illinoisstate.edu/actuary/ExamC/ExamCMay2005.pdf.
I understand all other parts except the first line of the answer key. For the geometric distribution, $Pr(X_1=2|\beta) = \frac{\beta^2}{(1+\beta)^3}$. I can see that it is based on the equation $\frac{\beta^{k-1}}{(1+β)^k}$, but I have no idea where did that come from. Can someone please help me? From what I understand, the geometric distribution is $p(1-p)^k$.
We have $\beta$, the expected value follows:$$\beta = \frac{1-p}p$$
Hence $\frac1p=\beta+1$ and $p=\frac1{\beta+1}$ $$(1-p)^2 p = \left( 1-\frac{1}{1+\beta}\right)^2\frac1{\beta+1}=\frac{\beta^2}{(1+\beta)^3}$$