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Let's break down the exercise step by step:

### Given Information:
For all \( n \in \mathbb{N}^* \) (the set of positive natural numbers), we define:

\[
S_n = 1 + 2 + \dots + n
\]

This is the sum of the first \( n \) natural numbers.

### Part 1: Calculating \( S_1 \), \( S_2 \), and \( S_3 \)

1. \( S_1 = 1 \)
2. \( S_2 = 1 + 2 = 3 \)
3. \( S_3 = 1 + 2 + 3 = 6 \)

### Part 2: Prove that for all \( n \in \mathbb{N}^* \):

\[
1 + 2 + \dots + (n-1) + n = \frac{n(n+1)}{2}
\]

**Proof using induction:**

#### Base Case: \( n = 1 \)
\[
S_1 = 1 = \frac{1(1+1)}{2} = \frac{2}{2} = 1
\]
This holds true.

#### Inductive Step:
Assume the formula holds for \( n = k \), i.e.,
\[
S_k = 1 + 2 + \dots + k = \frac{k(k+1)}{2}
\]
Now, prove for \( n = k+1 \):
\[
S_{k+1} = S_k + (k+1) = \frac{k(k+1)}{2} + (k+1)
\]
Factor the right-hand side:
\[
S_{k+1} = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}
\]
Thus, the formula holds for \( k+1 \), completing the inductive step.

By mathematical induction, the formula is proven for all \( n \in \mathbb{N}^* \).

### Part 3: Deduce that \( n(n+1) \) is even

Since \( n(n+1) \) is the product of two consecutive integers, one of them is always even. Therefore, \( n(n+1) \) is always even.

### Part 4: Calculate \( S_{100} \)

Using the formula for \( S_n \):
\[
S_{100} = \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = \frac{10100}{2} = 5050
\]

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Would you like more details or further explanations?

Here are some related questions to deepen your understanding:
1. How would you calculate the sum of the first \( n \) odd numbers?
2. What happens if you add the first \( n \) even numbers instead?
3. Can you generalize the formula for the sum of the squares of the first \( n \) numbers?
4. How does the concept of arithmetic progression relate to this sum formula?
5. How would the formula change if you were to sum from \( m \) to \( n \) instead of from 1?

**Tip:** When working with sums of sequences, understanding basic patterns and progressions can help simplify calculations.

For all n ∈ N*, we define: S_n = 1 + 2 + … + n. 1) Calculate S_1, S_2, and S_3. 2) Prove that for all n ∈ N*: 1 + 2 + … + n = n(n+1)/2. 3) Deduce that n(n+1) is even. 4) Calculate S_100.

Learn how to calculate the summation of the first n natural numbers using the formula S_n = n(n+1)/2. This step-by-step exercise covers basic arithmetic series, mathematical induction, and shows how n(n+1) is always even. Also includes a calculation for S_100.

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