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 $S_n$: $S_{100} = \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = \frac{10100}{2} = 5050$

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.