I see that you've uploaded an image containing several exercises related to sequences. Let's go over each one carefully.

Exercise 1:

For each of the sequences, calculate $u_0$, $u_1$, $u_2$, and $u_3$:

1. $u_n = \frac{1}{n}$
2. $u_n = 2n + 1$
3. $\begin{cases} u_0 = -1 \\ u_{n+1} = 3u_n + 2 \end{cases}$

Solution for Exercise 1:

1. $u_n = \frac{1}{n}$

   • $u_0 = \frac{1}{0}$ (undefined because division by zero is not allowed)
   • $u_1 = \frac{1}{1} = 1$
   • $u_2 = \frac{1}{2} = 0.5$
   • $u_3 = \frac{1}{3} \approx 0.33$

2. $u_n = 2n + 1$

   • $u_0 = 2(0) + 1 = 1$
   • $u_1 = 2(1) + 1 = 3$
   • $u_2 = 2(2) + 1 = 5$
   • $u_3 = 2(3) + 1 = 7$

3. $u_0 = -1$ and $u_{n+1} = 3u_n + 2$

   • $u_0 = -1$
   • $u_1 = 3(-1) + 2 = -3 + 2 = -1$
   • $u_2 = 3(-1) + 2 = -3 + 2 = -1$
   • $u_3 = 3(-1) + 2 = -3 + 2 = -1$

Exercise 2:

For each of the following sequences, express $u_n$ as a function of $u_{n-1}$:

1. $\begin{cases} u_0 = -2 \\ u_{n+1} = \frac{1}{2} u_n + 3 \end{cases}$
2. $\begin{cases} u_0 = 1 \\ u_{n+1} = nu_n + 3 \end{cases}$

Solution for Exercise 2:

1. First sequence:

   • Given: $u_0 = -2$ and $u_{n+1} = \frac{1}{2} u_n + 3$

   • To express $u_n$, we would recursively use the given formula.

   • $u_1 = \frac{1}{2}(-2) + 3 = -1 + 3 = 2$

   • $u_2 = \frac{1}{2}(2) + 3 = 1 + 3 = 4$

   • $u_3 = \frac{1}{2}(4) + 3 = 2 + 3 = 5$

   Therefore, the sequence starts with $-2, 2, 4, 5, \dots$.

2. Second sequence:

   • Given: $u_0 = 1$ and $u_{n+1} = nu_n + 3$

   • $u_1 = 0(1) + 3 = 3$

   • $u_2 = 1(3) + 3 = 3 + 3 = 6$

   • $u_3 = 2(6) + 3 = 12 + 3 = 15$

   Therefore, the sequence starts with $1, 3, 6, 15, \dots$.

Exercise 3:

In a library, an inventory is taken every year. The 2015 inventory shows a total of 8000 books. Each year, 10% of the books are discarded and 400 are added. The sequence $u_n$ represents the number of books in stock at the year $n$.

Solution for Exercise 3:

Let the number of books after year $n$ be represented by $u_n$. Initially, $u_0 = 8000$.

The recurrence relation for the number of books can be expressed as: $u_{n+1} = 0.9u_n + 400$

You can calculate the number of books for successive years by using this relation, starting with $u_0 = 8000$.

Let me know if you want the calculations for this exercise and any further details!

Do you have any questions on these solutions or would you like detailed steps?

Here are some related questions you might consider:

1. How do you express a recursive sequence as an explicit formula?
2. What are the differences between arithmetic and geometric sequences?
3. How do you calculate limits of sequences as $n \to \infty$?
4. How can you analyze the long-term behavior of a sequence like the one in Exercise 3?
5. What methods exist to solve non-linear recurrence relations?

Tip: In recursive sequences, it helps to compute several terms to identify a pattern before finding an explicit formula!