Let's go through these questions one by one:

1. Montrer que la somme de trois entiers consécutifs est divisible par 3.

We can represent the three consecutive integers as $n$, $n+1$, and $n+2$. The sum of these integers is:

$n + (n+1) + (n+2) = 3n + 3 = 3(n+1)$

Since $3(n+1)$ is divisible by 3, the sum of three consecutive integers is always divisible by 3.

2. Montrer de même que la somme de cinq entiers consécutifs est divisible par 5 et que la somme de sept entiers consécutifs est divisible par 7.

For five consecutive integers:

Represent the integers as $n$, $n+1$, $n+2$, $n+3$, and $n+4$. The sum is:

$n + (n+1) + (n+2) + (n+3) + (n+4) = 5n + 10 = 5(n+2)$

Thus, the sum is divisible by 5.

For seven consecutive integers:

Represent the integers as $n$, $n+1$, $n+2$, $n+3$, $n+4$, $n+5$, and $n+6$. The sum is:

$n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6) = 7n + 21 = 7(n+3)$

Thus, the sum is divisible by 7.

3. Montrer que la somme de six entiers consécutifs n'est jamais divisible par 6.

Represent the six consecutive integers as $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$. The sum is:

$n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15$

Since $6n + 15 = 6n + 3 \times 5$, the sum will not be divisible by 6 because of the remainder 15, which leaves 3 modulo 6.

4. Plus généralement, on étudie la somme de $k$ entiers consécutifs en notant $n$ le plus petit d'entre eux.

Let the integers be $n$, $n+1$, $n+2$, ..., $n+k-1$.

4a. Réduire cette somme à l'aide des suites arithmétiques.

The sum of the first $k$ consecutive integers starting at $n$ is:

$S_k = n + (n+1) + (n+2) + \dots + (n+k-1)$ $S_k = k \cdot n + \left(0 + 1 + 2 + \dots + (k-1)\right)$

The sum of the first $k-1$ integers $0 + 1 + 2 + \dots + (k-1)$ is $\frac{(k-1)k}{2}$, so:

$S_k = k \cdot n + \frac{k(k-1)}{2}$

4b. Montrer que cette somme est divisible par $k$ si et seulement si $k$ est impair.

The sum is:

$S_k = k \cdot n + \frac{k(k-1)}{2}$

For this to be divisible by $k$, the second term $\frac{k(k-1)}{2}$ must also be divisible by $k$. This happens if $k$ is odd, as $k-1$ is even, making $\frac{k-1}{2}$ an integer, which ensures the entire expression is divisible by $k$.

Would you like a deeper explanation for any part of this solution?

Here are five related questions:

1. How would the formula change for $k$-consecutive numbers starting from any non-integer?
2. What if the sequence skips values, e.g., taking only even integers?
3. How would we handle divisibility for a sequence of powers (e.g., squares or cubes)?
4. Is there a general rule for when sums of any sequence of integers are divisible by the count?
5. How can these sums be related to the properties of modular arithmetic?

Tip: When working with consecutive integers, grouping terms can help reveal patterns of divisibility.