I want to understand how this sum has been simplified for $n \ge 1$:
$$ \begin{split} \sum_{k=0}^n 3^k2^{n-k} &= 2^n\sum_{k=0}^n \left( \frac{3}{2} \right)^k \\ &= 2^n\frac{1-\left( \frac{3}{2} \right)^{n+1}}{1-\frac{3}{2}} \\ &=-2^{n+1} \left(1-\left( \frac{3}{2} \right)^{n+1}\right) \\ &= 3^{n+1} - 2^{n+1} \end{split} $$
I think that in the first step we have $2^n$ and $2^{-k}$ which is why we can write $2^n$ before the sum and $2^{-k} = \frac{1}{2^k}$. Then $3^k \cdot 2^{-k}$ would give $\frac{3^k}{2^k} = \big(\frac{3}{2}\big)^k$. Is that correct?
In the second step the getometric series has been used. But what happened in the last 2 steps?
It might be easier to see what's happening if you use variables $a$ and $b$ instead of the numbers $2$ and $3$:
$$\begin{align} \sum_{k=0}^na^kb^{n-k} &=b^n\sum_{k=0}^n(a/b)^k\\ &=b^n{1-(a/b)^{n+1}\over1-(a/b)}\\ &=b^{n+1}{1-(a/b)^{n+1}\over b-a}\\ &={b^{n+1}-a^{n+1}\over b-a} \end{align}$$
The crucial step here (after invoking the formula for the sum of a geometric series) involves multiplying numerator and denominator by $b$ so as which changes the denominator from $1-(a/b)$ to $b-a$. If we now substitute $a=3$ and $b=2$, we get
$$\sum_{k=0}^n3^k2^{n-k}={2^{n+1}-3^{n-1}\over2-3}={2^{n+1}-3^{n+1}\over-1}=3^{n+1}-2^{n+1}$$