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 $n \geq 0$.

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.