I want to understand how this sum has been simplified:
$$\sum_{k=2}^{n} \big( \frac{3}{2} \big)^{k-1} = \sum_{k=1}^{n-1} \big( \frac{3}{2} \big)^{k} = \sum_{k=0}^{n-1} \big( \frac{3}{2} \big)^{k}-1 $$ (Up until here I understand how the indexes have been shifted).
But how is this
$$= \frac{1-(\frac{3}{2})^n}{1-\frac{3}{2}} - 1 = -2 \cdot \big(1- \big( \frac{3}{2} \big)^n \big)- 1 = \frac{3^n}{2^{n-1}} -3$$
Can someone explain what happened in these last $3$ steps?
The first step is just the sum of a geometric series. We have $$1+a+a^2+\dots+a^{n-1}=\frac{a^n-1}{a-1}$$ That is easy to prove: suppose the sum is $S$. Then $aS$ is the same as $S$ except for the addition of the term $a^n$ and the deletion of the term 1, so subtracting we get the result.
The second step is that dividing by $1-\frac{3}{2}$ is the same as multiplying by $-2$.
The third step is just multiplying by $-2$ and noting that $-2-1=-3$.