Problem 22 - College Algebra with Modeling and Visualization

Question

Prove the statement by mathematical induction. $$ 3^{n}>2 n+1, \text { if } n \geq 2 $$

Step-by-Step Solution

Answer

The inequality \( 3^n > 2n + 1 \) is true for all \( n \geq 2 \).

Step 1: Base Case

First, we need to verify the base case for the smallest value of \( n \) greater than or equal to 2. We will start with \( n = 2 \). Substitute \( n = 2 \) into the inequality: \( 3^2 > 2 \times 2 + 1 \). Calculating each side yields \( 9 > 4 + 1 \), which simplifies to \( 9 > 5 \). The base case holds true.

Step 2: Inductive Hypothesis

For the inductive step, assume that the statement is true for \( n = k \), where \( k \geq 2 \). This means we assume: \( 3^k > 2k + 1 \). We will use this assumption to prove the statement for \( n = k + 1 \).

Step 3: Inductive Step

Now we need to prove that if the statement holds for \( n = k \), it also holds for \( n = k+1 \). We need to show that \( 3^{k+1} > 2(k+1) + 1 \). Start from the expression for \( 3^{k+1} \): \( 3^{k+1} = 3 \times 3^k \). Using the inductive hypothesis \( 3^k > 2k + 1 \), substitute to get \( 3 \times 3^k > 3 \times (2k + 1) \).

Step 4: Simplify Inductive Step

Simplify \( 3 \times (2k + 1) \) to get \( 3(2k + 1) = 6k + 3 \). Our goal is to show that \( 3(2k + 1) > 2(k+1) + 1 = 2k + 2 + 1 = 2k + 3 \). Thus, we need \( 6k + 3 > 2k + 3 \), which simplifies to \( 6k > 2k \). This is obviously true because \( 6k \) is always greater than \( 2k \) for \( k \geq 2 \).

Step 5: Conclusion

Since we've shown the inequality holds for the base case \( n=2 \) and that if it holds for \( n=k \) then it holds for \( n=k+1 \), by the principle of mathematical induction, the statement \( 3^n > 2n + 1 \) is true for all \( n \geq 2 \).

Key Concepts

Base Case

In mathematical induction, the base case is the foundation of the proof. It's the starting point which we verify first. For our problem, we need to show that the statement \( 3^n > 2n + 1 \) is true when \( n = 2 \), the smallest value in our specified range of \( n \).

Substituting \( n = 2 \) into the inequality gives us \( 3^2 > 2 \times 2 + 1 \). Calculate the left side to get 9, while the right side becomes 5. Since 9 is greater than 5, our base case is correct.

Once confirmed, this step reassures us that the statement holds at the beginning of our proof, forming a reliable base to proceed with the rest of the induction steps.

Inductive Hypothesis

The inductive hypothesis is a strategic assumption made to help carry out the induction proof. Suppose the statement is true for an arbitrary integer \( n = k \).

In our case, we assume:

• \( 3^k > 2k + 1 \)

This is our hypothesis and we don't prove this step; we take it as true help us in the next part of the induction proof.

The task is now to use this assumption for \( n = k \) and prove it for \( n = k + 1 \). The hypothesis acts as a bridge between the initial condition (base case) and the subsequent values.

Inductive Step

The inductive step involves showing that if the statement is true for \( n = k \), it must also be true for \( n = k + 1 \). This step is crucial because it illustrates the domino effect in induction.

Start with the expression \( 3^{k+1} = 3 \times 3^k \) and apply the inductive hypothesis. This implies \( 3^{k+1} > 3 \times (2k + 1) \). Simplify it:

• \( 3 \times (2k + 1) = 6k + 3 \)

Next, compare to \( 2(k+1) + 1 = 2k + 3 \).

Prove \( 6k + 3 > 2k + 3 \). On simplifying, it becomes clear that \( 6k > 2k \). Since this is true for \( k \geq 2 \), the inductive step is verified, showing the progression is maintained.

Inequalities

Inequalities are mathematical expressions indicating one quantity is larger or smaller than another. They are an integral part of our induction proof.

In the exercise, we are tasked with proving \( 3^n > 2n + 1 \) for \( n \geq 2 \). The crucial part of the proof revolves around ensuring these inequalities hold true through each step: from the base case, through assuming our inductive hypothesis, and verifying it at \( n = k + 1 \).

By effectively handling inequalities, we can layout a logical sequence that demonstrates how the initial assumption translates through every stage of induction, guaranteeing the statement holds for all applicable integers \( n \). It's through understanding and applying these inequalities that we can ensure the validity of induction principles.