The problem asks you to prove by mathematical induction the following formula for the sum of a geometric series:

$S_n = a + ar + ar^2 + \cdots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$

Step 1: Base Case (n = 1)

For $n = 1$, the left-hand side (LHS) of the equation is:

$S_1 = a$

The right-hand side (RHS) of the equation when $n = 1$ is:

$\frac{a(r^1 - 1)}{r - 1} = \frac{a(r - 1)}{r - 1} = a$

Thus, for $n = 1$, both sides are equal, so the base case holds.

Step 2: Inductive Hypothesis

Assume the formula is true for $n = k$, i.e.,

$S_k = a + ar + ar^2 + \cdots + ar^{k-1} = \frac{a(r^k - 1)}{r - 1}$

Step 3: Inductive Step

We need to prove the formula holds for $n = k+1$. The sum for $n = k+1$ is:

$S_{k+1} = a + ar + ar^2 + \cdots + ar^{k-1} + ar^k$

Using the inductive hypothesis for $S_k$, we can write:

$S_{k+1} = S_k + ar^k = \frac{a(r^k - 1)}{r - 1} + ar^k$

Now, factor out $ar^k$ on the right-hand side:

$S_{k+1} = \frac{a(r^k - 1)}{r - 1} + \frac{ar^k(r - 1)}{r - 1}$

Simplify the expression:

$S_{k+1} = \frac{a(r^k - 1 + r^k(r - 1))}{r - 1}$

Distribute $r^k$ over $(r - 1)$:

$S_{k+1} = \frac{a(r^k - 1 + r^{k+1} - r^k)}{r - 1} = \frac{a(r^{k+1} - 1)}{r - 1}$

This proves that the formula holds for $n = k+1$, completing the inductive step.

Conclusion:

By mathematical induction, the formula

$a + ar + ar^2 + \cdots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$

holds for all positive integers $n$.

Would you like any further clarification or details?

Related Questions:

1. How can the formula for the sum of an infinite geometric series be derived?
2. How does this formula change if $r = 1$?
3. Can you explain why the base case is important in mathematical induction?
4. How can we use mathematical induction to prove inequalities?
5. What are some other examples of geometric series in real-world applications?

Tip:

Always verify your base case carefully in induction problems, as it forms the foundation for the entire proof!