1. What is wrong with this \"proof\"? Prove that n! > 2 n. This problem is similar to Exercise 17 in Section 6.1 of your SNHU MAT299 textbook. Assume that this is true for n = k. In other words, k! > 2 k. Now (k + 1)! = (k + 1) k! > (k + 1) 2k ... by the induction hypothesis 2 * 2k... since k 1, or k + 1 2 = 2k+1. Thus this is true for n = k + 1 and we have proven the theorem. 2. Let n with n 1. Prove that 1 + x + x 2 + ... + xn = (1 - xn+1) / (1 - x). This problem is similar to Example 6.1.1 and to Exercise 7 in Section 6.1 of your SNHU MAT299 textbook. 3. Let n with n 2. Prove that 3n n 2n. Hint: If n 2, then 3n 2n + 2 (why?). This problem is similar to Example 6.1.3 and to Exercise 14 in Section 6.1 of your SNHU MAT299 textbook. 4. Let n . Prove that 3 | (4n + 7n + 1). Hint: You can write 7n+1 as (6+1) (7n). This problem is similar to Example 6.1.2 and to Exercise 11 in Section 6.1 of your SNHU MAT299 textbook. 5. Let n with n 2. Using the multiplication rule for derivatives ... (fg)' = f'g + fg' ... prove that the derivative of xn is nxn-1. This problem should be solved using mathematical induction in Section 6.1 of your SNHU MAT299 textbook. 6. Let n with n 2. If A1, A2, ..., An are sets, prove that A1 A2 ... An = An An-1 ... A1 where the unions are performed in order from left to right. This problem should be solved using mathematical induction in Section 6.1 of your SNHU MAT299 textbook. Three steps to solve a problem using induction: 1. Prove the statement to be true for n = 1 or 2 2. Assume the statement to be true for n = k. 3. Prove the statement to be true for n = k+1. Proof of Question 2. Step 1: For n =1, LHS = 1+x RHS = (1-x^2)/(1-x) = 1+x LHS = RHS Step 2: Assume it is true for n = k. 1 + x + ..... + x^k = (1-x^(k+1))/(1-x) Step 3: Prove it for n = k+1: LHS = 1 + x + ...... + x^(k+1) RHS = (1 - x^(k+2))/(1-x) LHS = (1 + x + ...... + x^k) + (x^(k+1)) = (1 - x^(k+1))/(1-x) + (x^(k+1)) (FROM STEP 2) = (1 - x^(k+2))/(1-x) = RHS in Step 3. Hence Proved. 3. Step 1: For n = 2 3^2>2*2^2 9>8 This is true. Step 2: Assume it to be true for n = k 3^k >= k*2^k Step 3: Prove for n = k+1 LHS = 3^(k+1) RHS = (k+1)*2^(k+1) = (2k +2)*2^k Simplifying LHS, 3*3^k > 3*(k*2^k) (FROM STEP 2) Using given information: 3k > 2k + 2 3k(2^k) > (2k +2 )(2^k) this is RHS Therefore LHS >= RHS Hence proved. 4. Step 1: Prove for n = 1: 4 + 7 + 1 = 12 3 | 12 this is true. Step 2: Assume true for n = k 3 | (4^k + 7^k +1) Therefore we can write (4^k + 7^k +1) as 3i, where i is some integer. Step 3: Prove for n = k+1: (4^(k+1) + 7^(k+1) + 1) = 4*4^k + 7*7^k + 1 = 4(4^(k) + 7^(k) + 1) + 3*7^k - 3 = 4*3i + 3j , where j = 7^k - 1, that is an integer. As both terms have 3 as a factor in them, therefore it is true for n = k+1 Hence proved. 5. Step 1: Prove for n = 2: x^2 = x*x (x^2)' = x'x +xx' = 2x = 2*x(2-1) = 2x True. Step 2: Assume true for n = k: (x^k)' = k*x^(k-1) Step 3: (x^(k+1))' = ((x^k)(x))' = (x^k)'x + x'x^k From Step 2: (k*x^(k-1))x + 1*x^k = k*x^k + x^k = (k+1)x^k True for n = k+1. Hence Proved. 5. Step 1: Prove for n = 2: x^2 = x*x (x^2)' = x'x +xx' = 2x = 2*x(2-1) = 2x True. Step 2: Assume true for n = k: (x^k)' = k*x^(k-1) Step 3: (x^(k+1))' = ((x^k)(x))' = (x^k)'x + x'x^k From Step 2: (k*x^(k-1))x + 1*x^k = k*x^k + x^k = (k+1)x^k True for n = k+1. Hence Proved