We are given a sequence $(U_n)_{n \in N}$ defined by $U_0 = 7$ and $U_{n+1} = \frac{1}{2}(U_n + 5)$. We define $V_n = U_n + a$. The problem asks to: 1. Find the real number $a$ such that $(V_n)_{n \in N}$ is a geometric sequence.

1. Problem Description

We are given a sequence $(U_n)_{n \in N}$ defined by $U_0 = 7$ and $U_{n+1} = \frac{1}{2}(U_n + 5)$.

We define $V_n = U_n + a$.

The problem asks to:

1. Find the real number $a$ such that $(V_n)_{n \in N}$ is a geometric sequence.

2. Let $a = -5$.

a. Express $V_n$ and $U_n$ as a function of $n$.

b. Show that $(U_n)_{n \in N}$ is decreasing on $N$.

c. Show that $(U_n)_{n \in N}$ is bounded below on $N$.

3. Express in terms of $n$ the sums $S_n = V_0 + V_1 + ... + V_n$ and $S'_n = U_0 + U_1 + U_2 + ... + U_n$.

4. Study the convergence of the sequences $(S_n)_{n \in N}$ and $(S'_n)_{n \in N}$.

2. Solution Steps

1. To find $a$ such that $V_n = U_n + a$ is a geometric sequence, we need $\frac{V_{n+1}}{V_n}$ to be a constant.

$V_{n+1} = U_{n+1} + a = \frac{1}{2}(U_n + 5) + a$.

Then, $\frac{V_{n+1}}{V_n} = \frac{\frac{1}{2}(U_n + 5) + a}{U_n + a} = \frac{U_n + 5 + 2a}{2(U_n + a)}$.

For this to be a constant, the numerator must be a multiple of the denominator, so we want $U_n + 5 + 2a = c(U_n + a)$ for some constant $c$. This can be written as $U_n + 5 + 2a = cU_n + ca$, so we have $c=1$ and $5 + 2a = a$. Thus $a = -5$.

If $a = -5$, then $V_n = U_n - 5$.

$V_{n+1} = U_{n+1} - 5 = \frac{1}{2}(U_n + 5) - 5 = \frac{1}{2}U_n + \frac{5}{2} - 5 = \frac{1}{2}U_n - \frac{5}{2} = \frac{1}{2}(U_n - 5) = \frac{1}{2}V_n$.

Thus, $\frac{V_{n+1}}{V_n} = \frac{1}{2}$, which is a constant. So $V_n$ is a geometric sequence with ratio $\frac{1}{2}$ and $a = -5$.

2. We are given $a = -5$, so $V_n = U_n - 5$. We know $V_n$ is a geometric sequence with ratio $q = \frac{1}{2}$.

$V_0 = U_0 - 5 = 7 - 5 = 2$.

Therefore, $V_n = V_0 q^n = 2(\frac{1}{2})^n = 2^{1-n}$.

Since $V_n = U_n - 5$, we have $U_n = V_n + 5 = 2^{1-n} + 5$.

3.

$U_{n+1} - U_n = (2^{-n} + 5) - (2^{1-n} + 5) = 2^{-n} - 2^{1-n} = 2^{-n} - 2 \cdot 2^{-n} = -2^{-n} < 0$.

Since $U_{n+1} - U_n < 0$, $U_{n+1} < U_n$, so the sequence $(U_n)_{n \in N}$ is decreasing.

4.

Since $U_n = 2^{1-n} + 5$, we have $U_n > 5$ for all $n$. So the sequence $(U_n)_{n \in N}$ is bounded below by

5.

5. $S_n = V_0 + V_1 + ... + V_n$ is the sum of a geometric sequence.

$S_n = \sum_{k=0}^n V_k = \sum_{k=0}^n 2(\frac{1}{2})^k = 2 \sum_{k=0}^n (\frac{1}{2})^k = 2 \frac{1 - (\frac{1}{2})^{n+1}}{1 - \frac{1}{2}} = 2 \frac{1 - (\frac{1}{2})^{n+1}}{\frac{1}{2}} = 4(1 - (\frac{1}{2})^{n+1}) = 4 - 2^{2-(n+1)} = 4 - 2^{1-n}$.

6. $S'_n = U_0 + U_1 + ... + U_n = \sum_{k=0}^n U_k = \sum_{k=0}^n (V_k + 5) = \sum_{k=0}^n V_k + \sum_{k=0}^n 5 = S_n + 5(n+1) = 4 - 2^{1-n} + 5n + 5 = 9 - 2^{1-n} + 5n$.

7. As $n \to \infty$, $2^{1-n} \to 0$, so $S_n \to 4$. Since $S_n$ approaches a finite number, $S_n$ is convergent.

As $n \to \infty$, $5n \to \infty$, so $S'_n = 9 - 2^{1-n} + 5n \to \infty$. Since $S'_n$ goes to infinity, $S'_n$ is divergent.

3. Final Answer

1. $a = -5$

2. a. $V_n = 2^{1-n}$ and $U_n = 2^{1-n} + 5$

b. $(U_n)_{n \in N}$ is decreasing on $N$.

c. $(U_n)_{n \in N}$ is bounded below on $N$.

3. $S_n = 4 - 2^{1-n}$ and $S'_n = 9 - 2^{1-n} + 5n$

4. $(S_n)_{n \in N}$ is convergent and $(S'_n)_{n \in N}$ is divergent.