Prove that 3^n - 1 is divisible by 2 for all n 1.

Here are the solutions to the problems. Question 5: Prove that 3^n - 1 is divisible by 2 for all n 1. We will prove this by mathematical induction. Let P(n) be the statement "3^n - 1 is divisible by 2". Step 1: Base Case (n=1). For n=1, we have 3^1 - 1 = 3 - 1 = 2. Since 2 is divisible by 2, P(1) is true. Step 2: Inductive Hypothesis. Assume that P(k) is true for some positive integer k 1. This means 3^k - 1 is divisible by 2. So, we can write 3^k - 1 = 2m for some integer m. From this, 3^k = 2m + 1. Step 3: Inductive Step. We need to prove that P(k+1) is true, i.e., 3^k+1 - 1 is divisible by 2. Consider the expression for n=k+1: 3^k+1 - 1 Rewrite 3^k+1 as 3 · 3^k: = 3 · 3^k - 1 Substitute 3^k = 2m + 1 from the inductive hypothesis: = 3(2m + 1) - 1 Distribute the 3: = 6m + 3 - 1 = 6m + 2 Factor out 2: = 2(3m + 1) Since 3m+1 is an integer, 2(3m+1) is a multiple of 2. Therefore, 3^k+1 - 1 is divisible by 2. Step 4: Conclusion. By the principle of mathematical induction, P(n) is true for all integers n 1. Thus, 3^n - 1 is divisible by 2 for all n 1. Question 6: Prove that 2^n n+1 for all n 1. We will prove this by mathematical induction. Let P(n) be the statement "2^n n+1". Step 1: Base Case (n=1). For n=1, we have 2^1 = 2 and 1+1 = 2. Since 2 2, P(1) is true. Step 2: Inductive Hypothesis. Assume that P(k) is true for some positive integer k 1. This means 2^k k+1. Step 3: Inductive Step. We need to prove that P(k+1) is true, i.e., 2^k+1 (k+1)+1, which simplifies to 2^k+1 k+2. Consider the left side of the inequality for n=k+1: 2^k+1 Rewrite 2^k+1 as 2 · 2^k: = 2 · 2^k From the inductive hypothesis, we know 2^k k+1. Multiply both sides by 2: 2 · 2^k 2(k+1) 2^k+1 2k+2 We want to show 2^k+1 k+2. We have 2^k+1 2k+2. Now, we need to show that 2k+2 k+2. Subtract k+2 from both sides: (2k+2) - (k+2) = 2k+2-k-2 = k Since k 1, we know k 0. Therefore, 2k+2 k+2. Combining the inequalities, we have 2^k+1 2k+2 k+2. Thus, 2^k+1 k+2. Step 4: Conclusion. By the principle of mathematical induction, P(n) is true for all integers n 1. Thus, 2^n n+1 for all n 1. Question 7: Prove that 4^n - 1 is divisible by 3 for all n 1. We will prove this by mathematical induction. Let P(n) be the statement "4^n - 1 is divisible by 3". Step 1: Base Case (n=1). For n=1, we have 4^1 - 1 = 4 - 1 = 3. Since 3 is divisible by 3, P(1) is true. Step 2: Inductive Hypothesis. Assume that P(k) is true for some positive integer k 1. This means 4^k - 1 is divisible by 3. So, we can write 4^k - 1 = 3m for some integer m. From this, 4^k = 3m + 1. Step 3: Inductive Step. We need to prove that P(k+1) is true, i.e., 4^k+1 - 1 is divisible by 3. Consider the expression for n=k+1: 4^k+1 - 1 Rewrite 4^k+1 as 4 · 4^k: = 4 · 4^k - 1 Substitute 4^k = 3m + 1 from the inductive hypothesis: = 4(3m + 1) - 1 Distribute the 4: = 12m + 4 - 1 = 12m + 3 Factor out 3: = 3(4m + 1) Since 4m+1 is an integer, 3(4m+1) is a multiple of 3. Therefore, 4^k+1 - 1 is divisible by 3. Step 4: Conclusion. By the principle of mathematical induction, P(n) is true for all integers n 1. Thus, 4^n - 1 is divisible by 3 for all n 1. Question 8: Show by mathematical induction that 7^(2n+3) + 1 is a multiple of 8 for all positive integers n. We will prove this by mathematical induction. Let P(n) be the statement "7^(2n+3) + 1 is a multiple of 8". Step 1: Base Case (n=1). For n=1, we have: 7^(2(1)+3) + 1 = 7^2+3 + 1 = 7^5 + 1 Calculate 7^5: 7^5 = 16807 So, 7^5 + 1 = 16807 + 1 = 16808. To check if 16808 is a multiple of 8: (16808)/(8) = 2101 Since 16808 is divisible by 8, P(1) is true. Step 2: Inductive Hypothesis. Assume that P(k) is true for some positive integer k 1. This means 7^(2k+3) + 1 is a multiple of 8. So, we can write 7^(2k+3) + 1 = 8m for some integer m. From this, $7^(