Confusion Over Sum of Geometric Series On pg. 88 of A First Course in Probability, it says 
$$ P_i - P_1 = P_1[(q/p) + (q/p)^2 + \cdots + (q/p)^{i-1})] $$
Therefore: 
$$P_i = \frac{1 - (q/p)^i}{1 - q/p}P_1 $$
The series on the right in the first equality sums to $\frac{1 - (q/p)^i}{1 - (q/p)}$, so how does the first step simplify to the second step? Thank you. 
 A: Using the summation formula finite Geometric Series, $$\frac{P_i}{P_1}=1+\frac qp+\left(\frac qp\right)^2+\cdots+\left(\frac qp\right)^{i-1}$$
$$=\frac{1-\left(\frac qp\right)^i}{1-\frac qp}$$
A: With regard to the geometric series on the right hand side of your first equation, the first term is $\frac{q}{p}$, so that the series will sum to $\large\frac{q}{p}\frac{1-\left(\frac{q}{p}\right)^i}{1-\frac{q}{p}}$ and not $\large\frac{1-\left(\frac{q}{p}\right)^i}{1-\frac{q}{p}}$. 
If you rearrange the first equation, you will obtain
$$P_i - P_1 = P_1[(q/p) + (q/p)^2 + \cdots + (q/p)^{i-1})]\\\Rightarrow P_i = P_1\color{blue}{[1 +(q/p) + (q/p)^2 + \cdots + (q/p)^{i-1})]}$$
The expression in blue is a geometric series with first term $1$ and ratio $\frac{q}{p}$, so that
$$P_i = P_1\color{blue}{\frac{1-\left(\frac{q}{p}\right)^i}{1-\frac{q}{p}}}$$
A: No, it doesn't.  Let $r = q/p$.  Then the first formula can be written as:  $$P_i = P_1(1 + r + r^2 + \cdots + r^{i-1}),$$ after moving the term $P_1$ from the left to the right hand side.  Then the series on the RHS is equal to $$1 + r + r^2 + \cdots + r^{i-1} = \frac{1-r^i}{1-r}.$$  Without the $1$ term, the series would be $$r + r^2 + \cdots + r^{i-1} = r \frac{1-r^{i-1}}{1-r}.$$
