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In mathematical induction, the base case serves as the foundational step where our goal is to show the given statement holds true for the initial integer, often the smallest possible value of a sequence. This is typically when the variable, such as \( n \), equals a starting value like 1 or sometimes 0.

For the given exercise, we begin by considering \( n = 1 \). Plugging this into our expression \( 3^n - 1 \) gives us \( 3^1 - 1 = 2 \). We need to check the divisibility by 2, and since 2 divided by 2 gives us a clean integer, this confirms our base case.

The base case is crucial. Without proving it, the domino effect of induction cannot start, as subsequent steps depend on this initial verification. Once we confirm the base case holds, we can move forward to make an assumption about the next step.

The inductive hypothesis involves making an assumption that a statement is true for some arbitrary integer, often denoted as \( k \). This step is a crucial bridge between the base case and proving the general case.

In this example, we assume that the statement \( 3^k - 1 \) is divisible by 2 for \( n = k \); \( k \) being any arbitrary positive integer. Our task at this stage isn't to prove anything new, but rather to establish a firm ground on which we build the final proof.

Think of the inductive hypothesis as setting the stage for the final act, where it ensures that we've implicitly considered all potential scenarios leading up to \( k \), preparing us for confirming the pattern continues with the next integer.

The inductive step in a proof by induction is where the magic happens. Here, we take the assumption from our inductive hypothesis and prove that the statement holds for the next integer, \( n = k+1 \). Successfully completing this step shows that if the statement holds for \( n = k \), it also holds for \( n = k+1 \).

In our exercise, we need to prove that \( 3^{k+1}-1 \) is divisible by 2, assuming the hypothesis that \( 3^k - 1 \) is divisible by 2. We begin by rewriting \( 3^{k+1} - 1 \) as \( 3 \times 3^k - 1 \). By reorganizing this using basic algebra, we find it is \( 3^k \times 2 + (3^k - 1) \).

The term \( 3^k \times 2 \) is clearly divisible by 2, and by our hypothesis, \( 3^k - 1 \) is also divisible by 2. Thus, this confirms that \( 3^{k+1} - 1 \) is divisible by 2, completing our proof. This step creates a chain reaction ensuring the statement is true for all natural numbers starting from the base case.