We are given two sequences $(U_n)$ and $(V_n)$ defined by the following relations: $U_0 = -\frac{3}{2}$ $U_{n+1} = \frac{2}{3}U_n - 1$ $V_n = 2U_n + a$ The questions are: 1. Determine the value of $a$ for which the sequence $(V_n)$ is a geometric sequence. Find the common ratio $q$ and the first term $V_0$.

1. Problem Description

We are given two sequences $(U_n)$ and $(V_n)$ defined by the following relations:

$U_0 = -\frac{3}{2}$

$U_{n+1} = \frac{2}{3}U_n - 1$

$V_n = 2U_n + a$

The questions are:

1. Determine the value of $a$ for which the sequence $(V_n)$ is a geometric sequence. Find the common ratio $q$ and the first term $V_0$.

2. Set $a=6$. Express $V_n$ and $U_n$ as functions of $n$.

3. Show that the sequence $(U_n)$ is decreasing.

4. Express the sums $S_n = V_0 + V_1 + V_2 + \dots + V_n$ and $S'_n = U_0 + U_1 + U_2 + \dots + U_n$ as functions of $n$.

5. Study the convergence of the sequence $(S'_n)$.

2. Solution Steps

1. For $(V_n)$ to be a geometric sequence, we must have $\frac{V_{n+1}}{V_n} = q$ for some constant $q$.

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

Since $V_n = 2U_n + a$, we have $U_n = \frac{V_n - a}{2}$.

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

For $(V_n)$ to be a geometric sequence, we must have $V_{n+1} = q V_n$ for some $q$. So $V_{n+1} = \frac{2}{3}V_n + \frac{1}{3}a - 2 = qV_n$. This implies $q = \frac{2}{3}$ and $\frac{1}{3}a - 2 = 0$.

Therefore, $\frac{1}{3}a = 2$, which gives $a = 6$.

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

The first term $V_0 = 3$, and the common ratio $q = \frac{2}{3}$.

2. With $a = 6$, $V_n = 2U_n + 6$. Since $(V_n)$ is geometric with $V_0 = 3$ and $q = \frac{2}{3}$, we have $V_n = V_0 q^n = 3(\frac{2}{3})^n$.

Then $2U_n + 6 = 3(\frac{2}{3})^n$, so $2U_n = 3(\frac{2}{3})^n - 6$, and $U_n = \frac{3}{2}(\frac{2}{3})^n - 3$.

3. To show that $(U_n)$ is decreasing, we need to show that $U_{n+1} - U_n < 0$.

$U_{n+1} - U_n = (\frac{3}{2}(\frac{2}{3})^{n+1} - 3) - (\frac{3}{2}(\frac{2}{3})^n - 3) = \frac{3}{2}(\frac{2}{3})^{n+1} - \frac{3}{2}(\frac{2}{3})^n = \frac{3}{2}(\frac{2}{3})^n(\frac{2}{3} - 1) = \frac{3}{2}(\frac{2}{3})^n(-\frac{1}{3}) = -\frac{1}{2}(\frac{2}{3})^n$.

Since $(\frac{2}{3})^n > 0$ for all $n$, we have $U_{n+1} - U_n = -\frac{1}{2}(\frac{2}{3})^n < 0$. Therefore, the sequence $(U_n)$ is decreasing.

4. $S_n = V_0 + V_1 + V_2 + \dots + V_n = \sum_{i=0}^n V_i$. Since $(V_n)$ is a geometric sequence, $S_n = V_0 \frac{1 - q^{n+1}}{1-q} = 3 \frac{1 - (\frac{2}{3})^{n+1}}{1 - \frac{2}{3}} = 3 \frac{1 - (\frac{2}{3})^{n+1}}{\frac{1}{3}} = 9 (1 - (\frac{2}{3})^{n+1}) = 9 - 9(\frac{2}{3})^{n+1}$.

$S'_n = U_0 + U_1 + U_2 + \dots + U_n = \sum_{i=0}^n U_i = \sum_{i=0}^n (\frac{3}{2}(\frac{2}{3})^i - 3) = \frac{3}{2} \sum_{i=0}^n (\frac{2}{3})^i - \sum_{i=0}^n 3 = \frac{3}{2} \frac{1 - (\frac{2}{3})^{n+1}}{1 - \frac{2}{3}} - 3(n+1) = \frac{3}{2} \frac{1 - (\frac{2}{3})^{n+1}}{\frac{1}{3}} - 3(n+1) = \frac{9}{2} (1 - (\frac{2}{3})^{n+1}) - 3(n+1) = \frac{9}{2} - \frac{9}{2}(\frac{2}{3})^{n+1} - 3n - 3 = \frac{3}{2} - \frac{9}{2}(\frac{2}{3})^{n+1} - 3n$.

5. We study the convergence of the sequence $(S'_n)$.

$S'_n = \frac{3}{2} - \frac{9}{2}(\frac{2}{3})^{n+1} - 3n$. As $n \to \infty$, $(\frac{2}{3})^{n+1} \to 0$, so $-\frac{9}{2}(\frac{2}{3})^{n+1} \to 0$. However, $-3n \to -\infty$.

Thus, $\lim_{n \to \infty} S'_n = \lim_{n \to \infty} (\frac{3}{2} - \frac{9}{2}(\frac{2}{3})^{n+1} - 3n) = -\infty$.

The sequence $(S'_n)$ diverges to $-\infty$.

3. Final Answer

1. $a = 6$, $q = \frac{2}{3}$, $V_0 = 3$.

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

3. The sequence $(U_n)$ is decreasing.

4. $S_n = 9 - 9(\frac{2}{3})^{n+1}$, $S'_n = \frac{3}{2} - \frac{9}{2}(\frac{2}{3})^{n+1} - 3n$.

5. The sequence $(S'_n)$ diverges to $-\infty$.