Problem 30 - Precalculus with Limits

Question

In Exercises 25 - 30, prove the inequality for the indicated integer values of \( n \). \(3^{n}>n 2^{n}, \quad n \geq 1\)

Step-by-Step Solution

Answer

Yes, the inequality \( 3^{n} > n * 2^{n} \) is true for all integers \( n \geq 1 \) based on the Principle of Mathematical Induction.

Step 1: Verify Base Case

The base case here is \( n = 1 \). Let's prove the inequality for this base case: \(3^{1} > 1 * 2^{1}\).\nHence, \(3 > 2\) is true.

Step 2: Assume True for n=k

Now let's assume that the given inequality is true for \( n = k \). That is, \(3^{k} > k * 2^{k}\). This is our induction hypothesis.

Step 3: Prove for n=k+1

We need to prove the inequality for \( n = k + 1 \). So, we need to prove \(3^{k+1} > (k+1)2^{k+1}\). Now we can start demonstrating it. We can say, \(3^{k+1} = 3 * 3^{k}\) and according to our induction hypothesis, \(3^{k} > k * 2^{k}\), so we have: \(3^{k+1} = 3 * 3^{k} > 3 * k * 2^{k}\). Given that \( k \geq 1 \), thus \( 3 * k \geq 3 * 1 = 3 \), it's clear that \( 3 * k * 2^{k} \geq 3 * 2^{k} = 3* 2 * 2^{k-1} > (k+1) * 2 * 2^{k-1} = (k+1) * 2^{k+1}\). Thus, we proved that if the inequality is true for \( n=k \), then it is true for \( n = k+1 \)

Step 4: Conclusion

Based on the Principle of Mathematical Induction, it is determined that the inequality \( 3^{n} > n * 2^{n} \) is indeed true for all integers \( n \geq 1 \).

Key Concepts

Inequality Proof

To grasp the concept of inequality proof through mathematical induction, it's essential to comprehend that we're establishing the truth of an inequality for all integers within a defined range. In this particular exercise, we aim to prove that the inequality \(3^n > n \cdot 2^n\) holds for all integers \(n \geq 1\). Each inequality has two parts:

• The left side \((3^n)\)
• The right side \((n \cdot 2^n)\)

The goal is to show that the left side is greater than the right side for the values of \(n\) mentioned.
Throughout the process, careful estimations and algebraic manipulations are used to confirm that the inequality holds. By following mathematical induction, we can demonstrate not only a specific case but a general case applicable to all integers \(n\) within the intended range. This systematic approach ensures precision and accuracy in proving inequalities.

Base Case

The base case is the starting point in mathematical induction. This step involves verifying the inequality for the initial value in the given set of integers.
In our exercise, the base case is when \(n = 1\). We need to confirm that the inequality \(3^1 > 1 \cdot 2^1\) is true.

• Calculate the left side: \(3^1 = 3\)
• Calculate the right side: \(1 \cdot 2^1 = 2\)

Since \(3 > 2\), the base case holds true.
By establishing that the base case is true, we lay the groundwork for further steps in mathematical induction. Without a true base case, the induction process cannot proceed. It serves as the foundation upon which all future steps rely.

Inductive Step

The inductive step is a crucial part of proving an inequality by mathematical induction. It involves two main tasks:

• Assume the inequality holds for an arbitrary integer \(n = k\)
• Show that if it holds for \(n = k\), it must also hold for \(n = k + 1\)

This is known as the induction hypothesis.
Here, we assume \(3^k > k \cdot 2^k\) as our hypothesis. The next step is to prove \(3^{k+1} > (k+1) \cdot 2^{k+1}\). Start with expressing \(3^{k+1}\) as \(3 \cdot 3^k\).
By using our assumption \(3^k > k \cdot 2^k\), we get:

• \(3^{k+1} = 3 \cdot 3^k > 3 \cdot k \cdot 2^k\)
• Recognize that \(3 \cdot k \geq 3\), making \(3 \cdot k \cdot 2^k\) greater than \((k+1) \cdot 2^{k+1}\)

Thus, the induction step is satisfied. By proving this step, we ensure that if the inequality holds for one integer, it must also hold for the next, leading to the conclusion that the inequality is true for all integers \(n \geq 1\). This step solidifies the chain of reasoning in the induction process.