Finding the Variance of Random Variable $X$ in a Geometric Distribution Given $$P(x) = p^{x−1} (1 − p)$$ for $$x = 1, 2, 3, 4, 5, . . .$$ and $$E(X-1)(X-2) = \frac{2p^2}{(1-p)^2}   $$ 
1)Find the Variance of X. 

I know that finding Var(X) is pretty straightforward and just equals $E(X^2) - (E(X))^2$ but in this case I don't know how to isolate E(X) out from $E(X-1)(X-2) = \frac{2p^2}{(1-p)^2} $. I'm thinking since they give us P(x) I should use it, but still confused about that too.

2) Calculate E(aX) for some number a that is |a| < 1.
3) Use the answer from part (c) to calculate $E(e^{tX})$ for some number t. For which values of t
does this expectation not exist?
 A: $\begin{align}\mathsf E((X-1)(X-2)) ~=~& \mathsf E(X^2-3X+2) \\[1ex]  ~=~& \mathsf {Var}(X)+\mathsf E(X)^2-3\mathsf E(X)+2\\[2ex]\therefore \mathsf{Var}(X) ~=~& 2p^2(1-p)^{-2}-2-\mathsf E(X)^2+3\mathsf E(X)
\end{align}$
So, if you can find $\mathsf E(X)$ from $P(x)= p^{x-1}(1-p)$ you can find the variance.
Use mathguy's hint.

You know that $~\sum\limits_{k=0}^\infty p^k = \frac 1{1-p}~$ from your studies on Series (it is the Geometric Series; learn it, love it, and use it often).
Then by taking derivatives wrt $p$ (not $x$, okay !): $$\frac{\mathrm d~~}{\mathrm d~p}\left(p^0+\sum_{k=1}^\infty p^k\right) ~=~ \frac{\mathrm d~~}{\mathrm d~p}\frac{1}{1-p} \\ ~\sum_{k=1}^\infty k p^{k-1} ~=~ \dfrac{1}{(1-p)^2}$$
Now because $~P(x) = (1-p)p^{x-1}~$ , then $~\mathsf E(X) = (1-p)\sum\limits_{x=1}^\infty x p^{x-1}~$ so...
A: to calculate $E(X)$
\begin{eqnarray*}
E(X)&=&\sum_{x=1}^\infty xp^{x-1}(1-p)=(1-p)(1+2p+3p^2+...)\\
    &=&(1-p)[(1+p+p^2+...)+(p+p^2+...)+(p^2+...)+...]\\
    &=&(1-p)\left(\frac{1}{1-p}+\frac{p}{1-p}+\frac{p^2}{1-p}+...\right)\\
    &=&1+p+p^2+...=\frac{1}{1-p}
\end{eqnarray*}
I hope it is clear enough
