We are tasked with proving by mathematical induction that for all positive integers $n$, the following proposition holds:

$1 + 3 + 3^2 + \dots + 3^{n-1} = \frac{3^n - 1}{2}.$

Step 1: Base Case

For $n = 1$:

The left-hand side (LHS) of the equation is:

$1 = 1.$

The right-hand side (RHS) of the equation is:

$\frac{3^1 - 1}{2} = \frac{3 - 1}{2} = \frac{2}{2} = 1.$

Since the LHS equals the RHS, the base case holds for $n = 1$.

Step 2: Inductive Hypothesis

Assume that the proposition holds for some positive integer $n = k$. That is, assume:

$1 + 3 + 3^2 + \dots + 3^{k-1} = \frac{3^k - 1}{2}.$

Step 3: Inductive Step

We need to prove that the proposition holds for $n = k + 1$, i.e., we need to show:

$1 + 3 + 3^2 + \dots + 3^k = \frac{3^{k+1} - 1}{2}.$

Using the inductive hypothesis, the sum up to $3^{k-1}$ is $\frac{3^k - 1}{2}$. Now, add $3^k$ to both sides of the inductive hypothesis:

$1 + 3 + 3^2 + \dots + 3^{k-1} + 3^k = \frac{3^k - 1}{2} + 3^k.$

We now simplify the right-hand side:

$\frac{3^k - 1}{2} + 3^k = \frac{3^k - 1}{2} + \frac{2 \cdot 3^k}{2} = \frac{3^k - 1 + 2 \cdot 3^k}{2} = \frac{3 \cdot 3^k - 1}{2}.$

Since $3 \cdot 3^k = 3^{k+1}$, we have:

$1 + 3 + 3^2 + \dots + 3^k = \frac{3^{k+1} - 1}{2}.$

Thus, the proposition holds for $n = k + 1$.

Step 4: Conclusion

By the principle of mathematical induction, the proposition is true for all positive integers $n$.

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Would you like further details or have any questions?

Five related questions:

1. How does mathematical induction work in proving statements?
2. Can the same method be used to prove geometric series for other bases?
3. Why is the base case important in induction?
4. How does the inductive step differ from a regular algebraic proof?
5. What happens if the base case does not hold?

Tip:

When applying induction, always double-check both the base case and inductive step for errors. Each part must be correct for the proof to be valid.