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The exercise asks to demonstrate by induction that for any real number \( q \neq 1 \), the following identity holds:

\[
\sum_{k=0}^{n} q^k = 1 + q + q^2 + \cdots + q^n = \frac{1 - q^{n+1}}{1 - q}
\]

### **Proof by Induction**

#### Step 1: Base Case
For \( n = 0 \), the sum becomes:

\[
\sum_{k=0}^{0} q^k = q^0 = 1
\]

On the right-hand side, we have:

\[
\frac{1 - q^{0+1}}{1 - q} = \frac{1 - q}{1 - q} = 1
\]

So the base case holds for \( n = 0 \).

#### Step 2: Induction Hypothesis
Assume that the formula is true for some integer \( n = m \), i.e.,

\[
\sum_{k=0}^{m} q^k = \frac{1 - q^{m+1}}{1 - q}
\]

#### Step 3: Inductive Step
We need to prove that the formula holds for \( n = m + 1 \). Consider the sum for \( n = m + 1 \):

\[
\sum_{k=0}^{m+1} q^k = \left( \sum_{k=0}^{m} q^k \right) + q^{m+1}
\]

Using the induction hypothesis for the first part:

\[
\sum_{k=0}^{m+1} q^k = \frac{1 - q^{m+1}}{1 - q} + q^{m+1}
\]

Now, simplify the expression:

\[
\frac{1 - q^{m+1}}{1 - q} + q^{m+1} = \frac{1 - q^{m+1} + q^{m+1}(1 - q)}{1 - q}
\]

\[
= \frac{1 - q^{m+2}}{1 - q}
\]

Thus, the formula holds for \( n = m+1 \).

#### Conclusion:
By the principle of mathematical induction, the formula is true for all \( n \geq 0 \).

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Would you like a further breakdown of the steps or details on the properties used here?

### Related Questions:
1. What is the significance of the assumption \( q \neq 1 \)?
2. Can this sum be extended to infinite series for \( |q| < 1 \)?
3. How does the formula change if \( q = 1 \)?
4. What are some applications of geometric series in real-life problems?
5. How can you derive the formula for the sum of a geometric progression without induction?

### Tip:
When proving by induction, ensure that the base case is correctly verified, as it forms the foundation for all subsequent steps.

Démontrer par récurrence que pour tout réel q ≠ 1 on a: ∑ (k=0 to n) q^k = 1 + q + q^2 + ... + q^n = (1 - q^(n+1)) / (1 - q)

Demonstrate using mathematical induction that for any real number q ≠ 1, the sum of the geometric series ∑ (k=0 to n) q^k equals (1 - q^(n+1)) / (1 - q). The solution covers concepts in series and induction, suitable for advanced high school or university-level students.

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