Mathematical Induction: Proving \(3^n > n \cdot 2^n\) for \(n > 1\)
Mathematical induction is a powerful proof technique that allows us to establish the truth of a statement for all natural numbers. In this article, we will use mathematical induction to prove the inequality \(3^n > n \cdot 2^n\) for all integers \(n\) greater than 1. To start the proof, we first need to establish the base case. Let's consider \(n = 2\). Plugging this value into the inequality, we have \(3^2 > 2 \cdot 2^2\), which simplifies to \(9 > 8\). Since this is true, we have successfully proven the base case. Next, we assume that the inequality holds for some arbitrary integer \(k\), i.e., \(3^k > k \cdot 2^k\). This assumption is known as the induction hypothesis. Our goal is to prove that the inequality also holds for \(k + 1\), i.e., \(3^{k+1} > (k+1) \cdot 2^{k+1}\). Using our induction hypothesis, we can rewrite the left side of the inequality as \(3 \cdot 3^k\) and the right side as \((k+1) \cdot 2 \cdot 2^k\). Now, our task is to show that \(3 \cdot 3^k > (k+1) \cdot 2 \cdot 2^k\). Expanding both sides, we have \(3 \cdot 3^k = 3^{k+1}\) and \((k+1) \cdot 2 \cdot 2^k = 2 \cdot (k+1) \cdot 2^k = (k+1) \cdot 2^{k+1}\). So, we need to prove that \(3^{k+1} > (k+1) \cdot 2^{k+1}\). By rearranging the terms, we can rewrite the inequality as \(\frac{3^{k+1}}{3^k} > \frac{(k+1) \cdot 2^{k+1}}{2^k}\), which simplifies to \(3 > \frac{k+1}{2}\). Now, we can see that for any positive integer \(k\), the left side of the inequality is always greater than 3, while the right side is always less than or equal to \(\frac{k+1}{2}\). Therefore, the inequality \(3^{k+1} > (k+1) \cdot 2^{k+1}\) holds true. Since we have proven the base case and shown that if the inequality holds for \(k\), it also holds for \(k+1\), we can conclude that \(3^n > n \cdot 2^n\) for all integers \(n\) greater than 1. In conclusion, mathematical induction is a powerful method that allows us to prove statements for all natural numbers. By using this technique, we have successfully proven the inequality \(3^n > n \cdot 2^n\) for all integers \(n\) greater than 1. This result has wide-ranging applications in various branches of mathematics and beyond.