Problem 16 - Precalculus with Limits

Question

In Exercises 11 - 24, use mathematical induction to prove the formula for every positive integer \( n \). \( 2\left(1 + 3 + 3^2 + 3^3 + \cdots + 3^{n - 1}\right) = 3^n - 1 \)

Step-by-Step Solution

Answer

By the principle of mathematical induction, the given formula \( 2\left(1 + 3 + 3^2 + 3^3 + \cdots + 3^{n - 1}\right) = 3^n - 1 \) can be proved to be valid for all positive integers \( n \).

Step 1: Test the Base Case

The base case will be tested for \( n = 1 \). Substituting \( n = 1 \) into the formula gives it as \( 2 \times 1 = 3^1 - 1 \). Solving both sides results in \( 2 = 2 \), which is true. This confirms the base case.

Step 2: Induction Hypothesis

Assume that the formula holds true for \( n = k \). This gives us: \( 2\left(1 + 3 + 3^2 + 3^3 + \cdots + 3^{k - 1}\right) = 3^k - 1 \). This is the induction hypothesis.

Step 3: Inductive step

We aim to show that it's true for \( n = k +1 \). Replacing \( n \) with \( k+1 \) in the statement results in: \( 2\left(1 + 3 + 3^2 + 3^3 + \cdots + 3^{k}\right) = 3^{k +1} - 1 \). Simplify left side it's: \( = 3^k - 1 + 2 \times 3^k \), and further simplify it's: \( = 3^k + 2 \times 3^k -1 = 3 \times 3^k -1 = 3^{k+1} - 1 \). The left-hand side of the equation indeed matches the right-hand side, so the formula holds true for \( n = k +1 \), given that it holds true for \( n = k \). Hence, by mathematical induction, the formula \( 2\left(1 + 3 + 3^2 + 3^3 + \cdots + 3^{n - 1}\right) = 3^n - 1 \) holds true for all positive integers \( n \).

Key Concepts

Proof by Induction

Proof by induction is a wonderful and powerful tool in mathematics used to demonstrate the validity of a statement. Especially when it's related to all natural numbers. The process consists of two main parts: the base case and the inductive step.
For example, if you’re proving a formula such as \(2(1 + 3 + 3^2 + \cdots + 3^{n-1}) = 3^n - 1\), you start by verifying it for an initial value, such as \(n=1\).

• Base Case: Confirm that the formula holds for the initial step.
• Inductive Step: Show that if the formula holds for an arbitrary \(n=k\), it holds for \(n=k+1\).

Together, these steps will help to demonstrate that the given statement is true for all integers above the base case, providing a "ladder" of truth onward from the base case.

Geometric Series

A geometric series is a series with a constant ratio between successive terms. This makes it relatively easy to derive a formula for the sum of the series.
The series can be written generally as:

• Terms: \(a, ar, ar^2, ar^3, \cdots, ar^{n-1}\)
• Sum: The sum of the first \(n\) terms is \(S_n = a\frac{1-r^n}{1-r}\) when \(r eq 1\).

In the exercise, the series given is \(1 + 3 + 3^2 + \cdots + 3^{n-1}\). Here, the first term \(a\) is 1, and the common ratio \(r\) is 3. By understanding geometric series, we see how the terms of this series contribute to the formula \(2(1 + 3 + 3^2 + \cdots + 3^{n-1}) = 3^n - 1\).

Base Case

The base case is the first step in a proof by induction, where you confirm that a statement holds true for the initial natural number, typically \(n = 1\).
This step is essential because it serves as the foundation for the entire inductive proof. It demonstrates that your formula starts off on the right track. Without it, there's no guarantee that induction will logically carry the statement forward.
In our exercise, testing the base case involves substituting \(n = 1\) into the formula \(2(1 + 3 + 3^2 + \cdots + 3^{n-1}) = 3^n - 1\) to confirm both sides are equal. Evaluation shows \(2 \times 1 = 3 - 1\), confirming the base case since both evaluate to 2.

Inductive Step

The inductive step is the core of proof by induction, bridging the gap between one stage of a reasoning process and the next.

• Assumption (Induction Hypothesis): Assume that the formula is true for a particular natural number \(n = k\).
• Demonstration: Then demonstrate that if the formula holds true for \(n = k\), it must also be true for \(n = k + 1\).

In the given exercise, after assuming the formula \(2(1 + 3 + 3^2 + \cdots + 3^{k-1}) = 3^k - 1\), you need to prove that \(2(1 + 3 + 3^2 + \cdots + 3^k) = 3^{k+1} - 1\).
By simplifying and replacing terms, you show both sides are equivalent, completing the inductive step. This step ensures that the truth cascades forward from any proven base.