Help in explaining this sigma notation breakdown

Question

I will appreciate some breakdown help which explains each step in the picture below to the last expression and the rules that applied to the changes. I am new to Sigma notations and thus confused.

Answer 1

(1)

$$\sum_{i=1}^{s}\color{red}{p}(1-p)^{i-1}=\color{red}{p}\sum_{i=1}^{s}(1-p)^{1-i}.$$

(2)

$$\begin{align}p\sum_{i=1}^{s}(1-p)^{i-1}&=p\{(1-p)^{1-1}+(1-p)^{2-1}+\cdots +(1-p)^{s-1}\}\\&=p\{(1-p)^0+(1-p)^1+\cdots+(1-p)^{s-1}\}\\&=p\sum_{i=0}^{s-1}(1-p)^i.\end{align}$$

(3)

Setting $S=\sum_{i=0}^{s-1}(1-p)^i$, then we have $$S=(1-p)^0+\color{red}{(1-p)^1+\cdots +(1-p)^{s-2}+(1-p)^{s-1}}$$ $$(1-p)S=\color{red}{(1-p)^1+(1-p)^2+\cdots +(1-p)^{s-1}}+(1-p)^s$$ So, substracting $(1-p)S$ from $S$ gives you $$\{1-(1-p)\}S=(1-p)^0-(1-p)^s\iff S=\sum_{i=0}^{s-1}(1-p)^i=\frac{1-(1-p)^s}{1-(1-p)}.$$

(4)

$$\begin{align}1-p\frac{1-(1-p)^s}{1-(1-p)}&=1-p\frac{1-(1-p)^s}{p}\\&=1-\{1-(1-p)^s\}\\&=(1-p)^s\end{align}$$

Answer 2

The sigma notation is a compact way of writing long, potentially infinite sums. For example, $\displaystyle{\sum_{i=1}^5}i^2 = 1+2^2+3^2+4^2+5^2$.

In your case, we have $\displaystyle{\sum_{i=1}^s}p(1-p)^{i-1} = p(1-p)^0 + p(1-p)^1 + p(1-p)^2 +...+p(1-p)^{s-1}$

which of course, is the same as $\displaystyle{\sum_{i=0}^{s-1}}p(1-p)^{i}$ when you expand this sum. This process is called reindexing. Factorising out the first $p$ gives you the second line $p\displaystyle{\sum_{i=0}^{s-1}}(1-p)^{i}$.

Now notice that $\displaystyle{\sum_{i=0}^{s-1}}(1-p)^{i}$ is a geometric series with initial term $1$ and constant multiple $(1-p)$. You use the formula for the sum of geometric series to get the third line.