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Mathematical Investigations IV Name: Mathematical Induction - Let the Dominoes Fall Prove each statement using mathematical induction. 1. Prove that for all positive integers, n, 1 1 1 n = ... 2n 1 1 3 3 5 (2n 1)(2n 1) Proof(by induction): Base Case: If n = 1, then LHS = 1 1 1 1 and RHS = 1 3 3 2 1 1 3 Inductive Step: suppose that for some value of n, 1 1 1 n = ... 2n 1 1 3 3 5 (2n 1)(2n 1) Then, 1 1 1 1 n 1 ... 1 3 3 5 (2n 1)(2n 1) (2n 1)(2n 3) 2n 1 (2n 1)(2n 3) n(2n 3) 1 (2n 1)(2n 3) (2n 1)(2n 3) 2n 2 3n 1 (2n 1)(2n 3) (2n 1)(n 1) (2n 1)(2n 3) (n 1) (2(n 1) 1) So for every natural number 1 1 1 n = ... 2n 1 1 3 3 5 (2n 1)(2n 1) Mathematical Investigations IV 2. Name: Prove that for all positive integers, n, 1 3 2 4 3 5 n ( n 2) n(n 1)(2n 7) 6 Proof (by induction): Base Case (n = 1) If n = 1, LHS = 1 3 3 and RHS = n(n 1)(2n 7) 1 2 9 3. 6 6 Inductive Step: Suppose that for some integer n, 1 3 2 4 3 5 n ( n 2) n(n 1)(2n 7) . 6 Then, 1 3 2 4 3 5 n (n 2) (n 1)(n 3) n(n 1)(2n 7) (n 1)(n 3) 6 n(2n 7) (n 1) (n 3) 6 2n 2 7n 6n 18 (n 1) 6 6 2n 2 13n 18 (n 1) 6 (2n 9)( n 4) (n 1) 6 (n 1)(n 2)(2(n 1) 7) 6 So, 1 3 2 4 3 5 n ( n 2) n(n 1)(2n 7) holds for all positive integers n. 6 Mathematical Investigations IV 3. Prove that Name: 22n1 1 is divisible by 3 for all positive integers, n. Proof by Induction: Base Case (n = 1) If n = 1, then 22 n1 1 23 1 9 , which is divisible by 3. Therefor the theorem holds when n = 1. Inductive Step: Suppose that 22k 1 1 is divisible by 3 for some positive integer, k. Then 22 k 1 1 3m for some positive integer m. It follows that 22 k 1 3m 1 Then, 22( k 1) 1 1 22 k 3 1 22 k 1 22 1 4 22 k 1 1 4(3m 1) 1 12m 4 1 12m 3 3 (4m 1) Which is clearly a multiple of 3. Therefore, by PMI, 22n1 1 is divisible by 3 for all positive integers, n. Mathematical Investigations IV 4. Prove that Name: n 1 for all positive integers, n. n 1 2 Proof (by induction): Base Case (n = 1) If n = 1, LHS = Inductive Step: Suppose that Then 1 1 1 which is clearly larger or equal to . 11 2 2 k 1 for some positive integer k. k 1 2 k 1 k 1 . (k 1) 1 k 2 k 1 k k 1 . , we will be done, since k 2 k 1 k 1 2 To that end, not that for positive integer k: If we can show that k 1 k (k 1)(k 1) k (k 2) k 2 k 1 (k 2)(k 1) (k 1)(k 2) k 2 2k 1 ( k 2 2k ) (k 2)(k 1) 1 (k 2)(k 1) 0 k 1 k k 1 k 1 0, it follows that . k 2 k 1 k 2 k 1 2 n 1 So by PMI, for all positive integers, n. n 1 2 Since Mathematical Investigations IV Name: