We are given a sequence $(U_n)$ defined by the initial term $U_0 = 1$ and the recursive relation $3U_{n+1} = U_n + 12$ for all $n \in N$. We also have $V_n = U_n - 6$. The questions ask us to: 1. Calculate the first three terms of the sequence $(U_n)$.

1. Problem Description

We are given a sequence $(U_n)$ defined by the initial term $U_0 = 1$ and the recursive relation $3U_{n+1} = U_n + 12$ for all $n \in N$. We also have $V_n = U_n - 6$. The questions ask us to:

1. Calculate the first three terms of the sequence $(U_n)$.

2. Show that $(V_n)$ is a geometric sequence and find its first term and common ratio.

3. Express $V_n$ and $U_n$ as functions of $n$.

4. Study the convergence of the sequences $(V_n)$ and $(U_n)$.

5. Calculate $S_n = V_0 + V_1 + V_2 + \dots + V_n$ and $S_n = U_0 + U_1 + U_2 + \dots + U_n$.

2. Solution Steps

1) Calculate the first three terms of the sequence $(U_n)$.

We have $U_0 = 1$.

Using the recursive relation $3U_{n+1} = U_n + 12$, we can find $U_1$ and $U_2$:

For $n=0$: $3U_1 = U_0 + 12 = 1 + 12 = 13$, so $U_1 = \frac{13}{3}$.

For $n=1$: $3U_2 = U_1 + 12 = \frac{13}{3} + 12 = \frac{13+36}{3} = \frac{49}{3}$, so $U_2 = \frac{49}{9}$.

Thus, the first three terms are $U_0 = 1$, $U_1 = \frac{13}{3}$, and $U_2 = \frac{49}{9}$.

2) Show that $(V_n)$ is a geometric sequence and find its first term and common ratio.

We have $V_n = U_n - 6$.

We want to show that $\frac{V_{n+1}}{V_n}$ is constant.

First, let's find $V_{n+1} = U_{n+1} - 6$.

From $3U_{n+1} = U_n + 12$, we have $U_{n+1} = \frac{1}{3}U_n + 4$.

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

Thus, $\frac{V_{n+1}}{V_n} = \frac{1}{3}$, which is constant. Therefore, $(V_n)$ is a geometric sequence with common ratio $q = \frac{1}{3}$.

The first term is $V_0 = U_0 - 6 = 1 - 6 = -5$.

3) Express $V_n$ and $U_n$ as functions of $n$.

Since $(V_n)$ is a geometric sequence with first term $V_0 = -5$ and common ratio $q = \frac{1}{3}$, we have

$V_n = V_0 \cdot q^n = -5 \cdot (\frac{1}{3})^n = -5 \cdot 3^{-n}$.

Since $V_n = U_n - 6$, we have $U_n = V_n + 6 = -5 \cdot 3^{-n} + 6$.

4) Study the convergence of the sequences $(V_n)$ and $(U_n)$.

Since $V_n = -5 \cdot 3^{-n}$, we have $\lim_{n \to \infty} V_n = \lim_{n \to \infty} -5 \cdot (\frac{1}{3})^n = -5 \cdot 0 = 0$. Thus, $(V_n)$ converges to

0. Since $U_n = -5 \cdot 3^{-n} + 6$, we have $\lim_{n \to \infty} U_n = \lim_{n \to \infty} (-5 \cdot (\frac{1}{3})^n + 6) = -5 \cdot 0 + 6 = 6$. Thus, $(U_n)$ converges to

6.

5) Calculate $S_n = V_0 + V_1 + V_2 + \dots + V_n$ and $S_n = U_0 + U_1 + U_2 + \dots + U_n$.

Since $V_n$ is a geometric sequence, the sum of the first $n+1$ terms is given by:

$S_n = \sum_{i=0}^n V_i = V_0 \cdot \frac{1 - q^{n+1}}{1 - q} = -5 \cdot \frac{1 - (\frac{1}{3})^{n+1}}{1 - \frac{1}{3}} = -5 \cdot \frac{1 - (\frac{1}{3})^{n+1}}{\frac{2}{3}} = -\frac{15}{2} (1 - (\frac{1}{3})^{n+1})$.

Let $S'_n = \sum_{i=0}^n U_i$. Since $U_i = V_i + 6$, we have

$S'_n = \sum_{i=0}^n (V_i + 6) = \sum_{i=0}^n V_i + \sum_{i=0}^n 6 = S_n + 6(n+1) = -\frac{15}{2} (1 - (\frac{1}{3})^{n+1}) + 6(n+1)$.

3. Final Answer

1) $U_0 = 1, U_1 = \frac{13}{3}, U_2 = \frac{49}{9}$

2) $V_n$ is a geometric sequence with first term $V_0 = -5$ and common ratio $q = \frac{1}{3}$.

3) $V_n = -5 \cdot 3^{-n}$ and $U_n = -5 \cdot 3^{-n} + 6$

4) $(V_n)$ converges to 0 and $(U_n)$ converges to

6. 5) $S_n = -\frac{15}{2} (1 - (\frac{1}{3})^{n+1})$ and $S'_n = -\frac{15}{2} (1 - (\frac{1}{3})^{n+1}) + 6(n+1)$.