Define the proposition P(n).

Okay David, let's tackle these induction problems. The document you sent provides the solutions, but I'll walk you through them step-by-step as if we were solving them from scratch. Problem 1: Show that 0^2 + 1^2 + 2^2 + ·s + n^2 = (n(n+1)(2n+1))/(6) for n 0. Step 1: Define the proposition P(n). Let P(n) be the statement 0^2 + 1^2 + 2^2 + ·s + n^2 = (n(n+1)(2n+1))/(6). Step 2: Base Case (n=0). We check if P(0) is true. The Left Hand Side (LHS) is 0^2 = 0. The Right Hand Side (RHS) is (0(0+1)(2(0)+1))/(6) = (0 · 1 · 1)/(6) = 0. Since LHS = RHS, P(0) is true. Step 3: Inductive Hypothesis. Assume P(k) is true for some integer k 0. That is, assume: 0^2 + 1^2 + 2^2 + ·s + k^2 = (k(k+1)(2k+1))/(6) Step 4: Inductive Step. We need to show that P(k+1) is true. This means we need to prove: 0^2 + 1^2 + 2^2 + ·s + k^2 + (k+1)^2 = ((k+1)((k+1)+1)(2(k+1)+1))/(6) 0^2 + 1^2 + 2^2 + ·s + k^2 + (k+1)^2 = ((k+1)(k+2)(2k+3))/(6) Start with the LHS of P(k+1): (0^2 + 1^2 + 2^2 + ·s + k^2) + (k+1)^2 Using the inductive hypothesis, substitute the sum up to k^2: (k(k+1)(2k+1))/(6) + (k+1)^2 Factor out the common term (k+1): (k+1) ( (k(2k+1))/(6) + (k+1) ) Find a common denominator for the terms inside the parenthesis: (k+1) ( (k(2k+1))/(6) + (6(k+1))/(6) ) (k+1) ( (2k^2+k + 6k+6)/(6) ) (k+1) ( (2k^2+7k+6)/(6) ) Factor the quadratic expression 2k^2+7k+6. It factors as (k+2)(2k+3). (k+1) ((k+2)(2k+3))/(6) ((k+1)(k+2)(2k+3))/(6) This matches the RHS of P(k+1). Step 5: Conclusion. Since P(0) is true and P(k) P(k+1), by the principle of mathematical induction, P(n) is true for all integers n 0. The statement is proven: 0^2 + 1^2 + 2^2 + ·s + n^2 = (n(n+1)(2n+1))/(6). --- Problem 2 (first part): Show that F_n-1F_n+1 = F_n^2 + (-1)^n for n 1. The Fibonacci numbers are defined by F_0 = 0, F_1 = 1, and F_k = F_k-1 + F_k-2 for k > 1. The first few Fibonacci numbers are F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, . Step 1: Define the proposition P(n). Let P(n) be the statement F_n-1F_n+1 = F_n^2 + (-1)^n. Step 2: Base Case (n=1). We check if P(1) is true. LHS: F_1-1F_1+1 = F_0F_2 = 0 · 1 = 0. RHS: F_1^2 + (-1)^1 = 1^2 - 1 = 1 - 1 = 0. Since LHS = RHS, P(1) is true. Step 3: Inductive Hypothesis. Assume P(k) is true for some integer k 1. That is, assume: F_k-1F_k+1 = F_k^2 + (-1)^k Step 4: Inductive Step. We need to show that P(k+1) is true. This means we need to prove: F_(k+1)-1F_(k+1)+1 = F_k+1^2 + (-1)^k+1 F_kF_k+2 = F_k+1^2 + (-1)^k+1 Start with the LHS of P(k+1): F_kF_k+2 Using the Fibonacci recurrence relation F_k+2 = F_k+1 + F_k: F_k(F_k+1 + F_k) F_kF_k+1 + F_k^2 From the inductive hypothesis, we know F_k^2 = F_k-1F_k+1 - (-1)^k. Substitute this into the expression: F_kF_k+1 + (F_k-1F_k+1 - (-1)^k) Factor out F_k+1: F_k+1(F_k + F_k-1) - (-1)^k Using the Fibonacci recurrence relation F_k + F_k-1 = F_k+1: F_k+1(F_k+1) - (-1)^k F_k+1^2 - (-1)^k Since -(-1)^k = (-1) · (-1)^k = (-1)^k+1: F_k+1^2 + (-1)^k+1 This matches the RHS of P(k+1). Step 5: Conclusion. Since P(1) is true and P(k) P(k+1), by the principle of mathematical induction, P(n) is true for all integers n 1. The statement is proven: F_n-1F_n+1 = F_n^2 + (-1)^n. That's 2 down. 3 left today — send the next one.