Show by mathematical induction that 7^(2n+3) + 1 is a multiple of 8 for all positive integers n.

Here are the solutions to problems 8 and 9. 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^(2k+3) = 8m - 1. Step 3: Inductive Step. We need to prove that P(k+1) is true, i.e., 7^(2(k+1)+3) + 1 is a multiple of 8. First, simplify the exponent for n=k+1: 2(k+1)+3 = 2k+2+3 = 2k+5 So we need to show 7^(2k+5) + 1 is a multiple of 8. Consider the expression for n=k+1: 7^(2k+5) + 1 Rewrite 7^(2k+5) as 7^2 · 7^(2k+3): = 7^2 · 7^(2k+3) + 1 = 49 · 7^(2k+3) + 1 Substitute 7^(2k+3) = 8m - 1 from the inductive hypothesis: = 49(8m - 1) + 1 Distribute the 49: = 49 · 8m - 49 + 1 = 49 · 8m - 48 Factor out 8: = 8(49m - 6) Since 49m - 6 is an integer, 8(49m - 6) is a multiple of 8. Therefore, 7^(2(k+1)+3) + 1 is a multiple of 8. Step 4: Conclusion. By the principle of mathematical induction, P(n) is true for all positive integers n. Thus, 7^(2n+3) + 1 is a multiple of 8 for all positive integers n. Question 9: Given u_n+1 = 3u_n + 4, u_1 = 1. Prove u_n = 3^n - 2. We will prove this by mathematical induction. Let P(n) be the statement "u_n = 3^n - 2". Step 1: Base Case (n=1). For n=1, we check if P(1) is true. Given u_1 = 1. Using the proposed formula u_n = 3^n - 2: u_1 = 3^1 - 2 = 3 - 2 = 1 Since the value from the formula matches the given initial condition (u_1 = 1), P(1) is true. Step 2: Inductive Hypothesis. Assume that P(k) is true for some positive integer k 1. This means we assume u_k = 3^k - 2. Step 3: Inductive Step. We need to prove that P(k+1) is true, i.e., u_k+1 = 3^k+1 - 2. From the given recurrence relation, we know: u_k+1 = 3u_k + 4 Substitute the inductive hypothesis u_k = 3^k - 2 into this equation: u_k+1 = 3(3^k - 2) + 4 Distribute the 3: u_k+1 = 3 · 3^k - 3 · 2 + 4 u_k+1 = 3^k+1 - 6 + 4 u_k+1 = 3^k+1 - 2 This is exactly the formula for u_k+1 that we wanted to prove. Thus, P(k+1) is true. Step 4: Conclusion. By the principle of mathematical induction, the statement u_n = 3^n - 2 is true for all positive integers n 1.