We are given two sequences, $(U_n)_{n \in \mathbb{N}}$ and $(V_n)_{n \in \mathbb{N}}$, defined by $U_0 = -\frac{3}{2}$, $U_{n+1} = \frac{2}{3} U_n - 1$, and $V_n = 2 U_n + 6$. We need to show that $(V_n)_{n \in \mathbb{N}}$ is a geometric sequence, and determine its common ratio $q$ and its first term $V_0$. Also, we need to express $V_n$ and $U_n$ in terms of $n$.

1. Problem Description

We are given two sequences, $(U_n)_{n \in \mathbb{N}}$ and $(V_n)_{n \in \mathbb{N}}$, defined by $U_0 = -\frac{3}{2}$, $U_{n+1} = \frac{2}{3} U_n - 1$, and $V_n = 2 U_n + 6$.

We need to show that $(V_n)_{n \in \mathbb{N}}$ is a geometric sequence, and determine its common ratio $q$ and its first term $V_0$. Also, we need to express $V_n$ and $U_n$ in terms of $n$.

2. Solution Steps

First, let's find the value of $V_0$:

$V_0 = 2U_0 + 6 = 2(-\frac{3}{2}) + 6 = -3 + 6 = 3$

Next, we want to find $V_{n+1}$ in terms of $V_n$.

$V_{n+1} = 2U_{n+1} + 6 = 2(\frac{2}{3}U_n - 1) + 6 = \frac{4}{3}U_n - 2 + 6 = \frac{4}{3}U_n + 4$

Now, we want to relate $V_{n+1}$ to $V_n$. Since $V_n = 2U_n + 6$, we have $U_n = \frac{V_n - 6}{2}$. Substitute this into the expression for $V_{n+1}$:

$V_{n+1} = \frac{4}{3}(\frac{V_n - 6}{2}) + 4 = \frac{2}{3}(V_n - 6) + 4 = \frac{2}{3}V_n - 4 + 4 = \frac{2}{3}V_n$

Since $V_{n+1} = \frac{2}{3}V_n$, the sequence $(V_n)$ is a geometric sequence with common ratio $q = \frac{2}{3}$ and first term $V_0 = 3$.

Now, we express $V_n$ in terms of $n$. Since it is a geometric sequence:

$V_n = V_0 \cdot q^n = 3 \cdot (\frac{2}{3})^n$

Finally, we express $U_n$ in terms of $n$.

Since $V_n = 2U_n + 6$, we have $U_n = \frac{V_n - 6}{2}$. Substituting the expression for $V_n$:

$U_n = \frac{3 (\frac{2}{3})^n - 6}{2} = \frac{3}{2}(\frac{2}{3})^n - 3$

3. Final Answer

The sequence $(V_n)$ is a geometric sequence with the first term $V_0 = 3$ and the common ratio $q = \frac{2}{3}$.

The expressions for $V_n$ and $U_n$ in terms of $n$ are:

$V_n = 3 (\frac{2}{3})^n$

$U_n = \frac{3}{2}(\frac{2}{3})^n - 3$