Problem 26 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

The inequality is proven true for all \( n \geq 2 \) by mathematical induction.

Step 1: Base Case

Verify the inequality for the smallest possible value of \( n \), which is \( n = 2 \). Substitute \( n = 2 \) into the inequality: \[ 3^2 > 2 \times 2 + 1 \]Calculate the left side and the right side:\[ 9 > 4 + 1 \]\[ 9 > 5 \]The inequality holds true for \( n = 2 \).

Step 2: Inductive Hypothesis

Assume the statement is true for some arbitrary integer \( k \geq 2 \). This gives us that:\[ 3^k > 2k + 1 \]This assumption will be used to prove the statement for \( k + 1 \).

Step 3: Inductive Step

Prove the inequality for \( n = k + 1 \) using the inductive hypothesis. We need to show:\[ 3^{k+1} > 2(k+1) + 1 \]Rewrite \( 3^{k+1} \) as \( 3 \times 3^k \) and use the inductive hypothesis:\[ 3^{k+1} = 3 \times 3^k > 3 \times (2k + 1) \]Compute the right-hand side:\[ 3 \times (2k + 1) = 6k + 3 \]We need to show:\[ 6k + 3 > 2k + 3 \]Simplifying, this becomes:\[ 4k > 0 \]Since \( k \geq 2 \), \( 4k > 0 \) is always true.

Step 4: Conclusion

Since the base case is true and the inductive step holds, by the principle of mathematical induction, the statement \( 3^n > 2n + 1 \) is true for all integers \( n \geq 2 \).

Key Concepts

Base Case

In mathematical induction, the base case is the starting point of our proof. It's like testing the waters to see if the initial condition holds true. For this exercise, our goal is to show that the inequality \(3^n > 2n + 1\) is true for all integers \(n \geq 2\).

We begin by checking the smallest possible value of \(n\), which is \(n = 2\). By substituting \(n = 2\) into our inequality, we get:

• Left Side: \(3^2 = 9\)
• Right Side: \(2 \times 2 + 1 = 5\)

Since \(9 > 5\), the inequality holds true at \(n = 2\). The base case serves as the foundation of your inductive proof, ensuring that the statement works at the start of our sequence.

Inductive Hypothesis

The inductive hypothesis is an assumption that the statement we're trying to prove holds true for a particular integer \(k\).

For this inequality proof, we assume that for some arbitrary \(k \geq 2\), the inequality \(3^k > 2k + 1\) is true. This assumption isn't just guessed; it serves as a stepping stone to bridge the gap to the next value in our sequence. The magic of the inductive hypothesis is its role in helping prove the statement for \(k + 1\). It's like setting a halfway point that allows us to incrementally prove the entire statement step by step.

Inductive Step

The inductive step is where we leverage the inductive hypothesis to demonstrate that if it holds for some integer \(k\), it must also be true for \(k + 1\).

To show this, we start by setting up the inequality for \(k + 1\):

• We need to prove \(3^{k+1} > 2(k+1) + 1\).
• Rewrite \(3^{k+1}\) as \(3 \times 3^k\), then use our inductive hypothesis: \(3 \times 3^k > 3 \times (2k + 1)\).

By performing calculations, such as simplifying \(3 \times (2k + 1) = 6k + 3\), and comparing this to \(2(k+1) + 1 = 2k + 3\), you end up with:

• Simplification shows \(4k > 0\).
• Since \(k \geq 2\), clearly \(4k > 0\).

This completes the inductive step, showing that the inequality holds for \(k + 1\) providing strength to our overall proof.

Inequality Proof

An inequality proof using mathematical induction requires us to manage both the structure of our logic and the algebra involved.

By understanding each component—base case, inductive hypothesis, and inductive step—we are able not only to correctly perform the algebra but also illustrate the logical flow. Specifically, in this exercise, knowing how to manipulate and simplify expressions like \(3 \times 3^k\) and comparing \(6k + 3\) with \(2k + 3\) allows us to confidently verify each inequality.

• The base case showed the inequality true for \(n=2\).
• The inductive hypothesis set the foundation for proving it true for all \(k\).
• The inductive step confirmed the statement for \(k+1\) using that foundation.

Taking these steps together, we establish the proof for all integers \(n \geq 2\), showing the power of mathematical induction when dealing with inequalities.