Mathematical Induction: Proving \( 3^{n}>n .2^{n} \)

Mathematical induction is a powerful tool in the field of mathematics that allows us to prove statements for an infinite number of cases. In this article, we will use mathematical induction to prove the inequality \( 3^{n}>n .2^{n} \) whenever \( n \) is an integer greater than 1. To begin, let's establish the base case for our proof. When \( n = 2 \), we have \( 3^{2} = 9 \) and \( 2 .2^{2} = 8 \). Clearly, \( 9 > 8 \), so the inequality holds true for \( n = 2 \). Next, we assume that the inequality holds true for some arbitrary integer \( k \), where \( k \) is greater than 1. That is, we assume \( 3^{k} > k .2^{k} \). Now, we need to prove that the inequality also holds true for \( n = k+1 \). We start by considering the left side of the inequality, \( 3^{k+1} \). Using the properties of exponents, we can rewrite this expression as \( 3 .3^{k} \). Similarly, we can rewrite the right side of the inequality, \( (k+1) .2^{k+1} \), as \( k .2^{k} + 2^{k+1} \). Now, let's compare the two expressions. We know that \( 3^{k} > k .2^{k} \) from our assumption. Therefore, we can replace \( 3^{k} \) in the expression \( 3 .3^{k} \) with \( k .2^{k} \). This gives us \( 3 .k .2^{k} \) on the left side of the inequality. On the right side, we have \( k .2^{k} + 2^{k+1} \). Now, let's simplify the right side of the inequality. We can factor out \( k .2^{k} \) from the expression \( k .2^{k} + 2^{k+1} \), giving us \( k .2^{k} \cdot (1 + 2) \), which simplifies to \( 3 .k .2^{k} \). Therefore, we have shown that \( 3^{k+1} > (k+1) .2^{k+1} \), which means the inequality holds true for \( n = k+1 \). By the principle of mathematical induction, we have shown that the inequality \( 3^{n} > n .2^{n} \) holds true for all integers greater than 1. In conclusion, using mathematical induction, we have proven the inequality \( 3^{n} > n .2^{n} \) for all integers greater than 1. This result demonstrates the power and versatility of mathematical induction in proving statements that hold true for an infinite number of cases.