Prove the following statements using the Principle of Mathematical Induction: 1. 3^n-1 is divisible by 2 for all n ext extgreater ext 1. 2. 2^n ext extgreater ext n+1 for all n ext extgreater ext 1.

Here are the solutions to the problems using the Principle of Mathematical Induction. 1. Prove that 3^n-1 is divisible by 2 for all n 1. Let P(n) be the statement "3^n-1 is divisible by 2". Base Case (n=1): P(1): 3^1-1 = 2. Since 2 is divisible by 2, P(1) is true. Inductive Hypothesis: Assume P(m) is true for some positive integer m. That is, assume 3^m-1 is divisible by 2. This means 3^m-1 = 2k for some integer k. So, 3^m = 2k+1. Inductive Step: We need to prove that P(m+1) is true, i.e., 3^m+1-1 is divisible by 2. Consider 3^m+1-1: 3^m+1-1 = 3 · 3^m - 1 Substitute 3^m = 2k+1 from the inductive hypothesis: 3(2k+1) - 1 6k + 3 - 1 6k + 2 2(3k+1) Since 2(3k+1) is a multiple of 2, 3^m+1-1 is divisible by 2. Thus, P(m+1) is true. By the Principle of Mathematical Induction, P(n) is true for all n 1. The proof is complete. --- 2. Prove that 2^n n+1 for all n 1. Let P(n) be the statement 2^n n+1. Base Case (n=1): LHS = 2^1 = 2 RHS = 1+1 = 2 Since 2 2, P(1) is true. Inductive Hypothesis: Assume P(m) is true for some positive integer m. That is, assume 2^m m+1. Inductive Step: We need to prove that P(m+1) is true, i.e., 2^m+1 (m+1)+1, which simplifies to 2^m+1 m+2. Consider the LHS of P(m+1): 2^m+1 = 2 · 2^m By the Inductive Hypothesis, we know 2^m m+1. Multiply both sides by 2: 2 · 2^m 2(m+1) 2^m+1 2m+2 We want to show 2^m+1 m+2. Since m 1, we know that m 0. Therefore, 2m+2 = m+m+2 m+0+2 = m+2. So, we have: 2^m+1 2m+2 m+2 Thus, 2^m+1 m+2. This proves P(m+1) is true. By the Principle of Mathematical Induction, P(n) is true for all n 1. The proof is complete.