The problem defines a sequence $(U_n)_{n \in \mathbb{N}}$ by $U_0 = 1$ and $U_{n+1} = \frac{2U_n}{U_n + 2}$. We are asked to: 1. Show that the sequence $(V_n)_{n \in \mathbb{N}}$ defined by $V_n = \frac{1}{U_n}$ is an arithmetic sequence.

1. Problem Description

The problem defines a sequence $(U_n)_{n \in \mathbb{N}}$ by $U_0 = 1$ and $U_{n+1} = \frac{2U_n}{U_n + 2}$.

We are asked to:

1. Show that the sequence $(V_n)_{n \in \mathbb{N}}$ defined by $V_n = \frac{1}{U_n}$ is an arithmetic sequence.

2. Find the common difference (reason) and the first term of $(V_n)$.

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

4. Calculate the sum $S_n = V_0 + V_1 + V_2 + ... + V_n$ as a function of $n$ using two different methods.

2. Solution Steps

1. To show that $V_n$ is an arithmetic sequence, we need to show that $V_{n+1} - V_n$ is constant.

$V_{n+1} = \frac{1}{U_{n+1}} = \frac{1}{\frac{2U_n}{U_n + 2}} = \frac{U_n + 2}{2U_n} = \frac{U_n}{2U_n} + \frac{2}{2U_n} = \frac{1}{2} + \frac{1}{U_n} = \frac{1}{2} + V_n$

Therefore, $V_{n+1} - V_n = \frac{1}{2}$. Since this difference is constant, $V_n$ is an arithmetic sequence.

2. The common difference (reason) is $\frac{1}{2}$.

The first term is $V_0 = \frac{1}{U_0} = \frac{1}{1} = 1$.

3. Since $V_n$ is an arithmetic sequence with first term $V_0 = 1$ and common difference $r = \frac{1}{2}$, we can write $V_n$ as:

$V_n = V_0 + nr = 1 + n(\frac{1}{2}) = 1 + \frac{n}{2} = \frac{n+2}{2}$.

Since $V_n = \frac{1}{U_n}$, we have $U_n = \frac{1}{V_n} = \frac{1}{\frac{n+2}{2}} = \frac{2}{n+2}$.

4. Method 1: Using the formula for the sum of an arithmetic sequence.

$S_n = V_0 + V_1 + ... + V_n = \sum_{i=0}^n V_i$.

The sum of an arithmetic sequence is given by:

$S_n = \frac{(V_0 + V_n)(n+1)}{2}$.

Substituting $V_0 = 1$ and $V_n = \frac{n+2}{2}$, we get

$S_n = \frac{(1 + \frac{n+2}{2})(n+1)}{2} = \frac{(\frac{2+n+2}{2})(n+1)}{2} = \frac{(\frac{n+4}{2})(n+1)}{2} = \frac{(n+4)(n+1)}{4} = \frac{n^2 + 5n + 4}{4}$.

Method 2: Using the formula $V_n = 1 + \frac{n}{2}$.

$S_n = \sum_{i=0}^n V_i = \sum_{i=0}^n (1 + \frac{i}{2}) = \sum_{i=0}^n 1 + \frac{1}{2}\sum_{i=0}^n i = (n+1) + \frac{1}{2} \frac{n(n+1)}{2} = (n+1) + \frac{n(n+1)}{4} = \frac{4(n+1) + n(n+1)}{4} = \frac{4n+4 + n^2 + n}{4} = \frac{n^2 + 5n + 4}{4}$.

3. Final Answer

1. $V_n$ is an arithmetic sequence because $V_{n+1} - V_n = \frac{1}{2}$.

2. The reason is $\frac{1}{2}$ and the first term is

1.

3. $V_n = \frac{n+2}{2}$ and $U_n = \frac{2}{n+2}$.

4. $S_n = \frac{n^2 + 5n + 4}{4}$.