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Math 433 Induction Practice Problem 1 Prove by induction that if A = {1, 2, 3, . . . , n}, then the power set, P(A), has 2n elements. Problem 2 Every integer greater than 1 is divisible by a prime. Problem 3 For every n ∈ N, 2n3 + 3n2 + n is divisible by 6. Problem 4 Prove by induction that 1 · 2 + 2 · 3 + 3 · 4 + · · · n · (n + 1) = n(n + 1)(n + 2) . 3 (1) Problem 5 If fn are the Fibonacci numbers, then for all n ∈ N, n X f2r−1 = f2n . (2) r=1 1. Solution. Recall that for a (particular) set, A, the power set, P(A), is the collection of subsets of A. We’ll use induction. What is the sequence of statements? (Sn): The number of elements in P({1, 2, . . . , n}) is 2n . First we check that (S1) is true. But, the subsets of the set {1} are {∅, {1}}. That is, P({1}) = {∅, {1}}. Hence, there are exactly two elements in P({1}). Now assume that (S n) is true. That is, assume that the number of elements in P({1, 2, . . . , n}) is 2n . We have to show that (S n+1) is true. So, we have to count the number of subsets of {1, 2, . . . , n, n + 1}. But, a subset of this last set is one of two types. Either the subset contains n + 1 or it doesn’t. Those subsets which do not contain n + 1 are just subsets of {1, 2, . . . , n}. And we know how many of these we have: 2n . Those subsets which contain n + 1 can be described by taking a subset of {1, 2, . . . , n} and throwing in the element n + 1. For example, if n + 1 = 5 and the subset of {1, 2, . . . , 5} which we would like to describe is {1, 3, 5}, we could say that {1, 3, 5} = {1, 3} ∪ {5}. Or, more generally, every set which contains n + 1 can be written as A ∪ {n + 1}, where A is a subset of {1, 2, . . . , n}. Hence, to count these sets we only have to count the A’s which are subsets of {1, 2, . . . , n}. Again, this is just 2n . So, we’ve divided our problem into counting in two distinct situations and in each case there are 2n elements. Therefore, the total number of elements is 2n + 2n = 2 · 2n = 2n+1 . Therefore, we’ve verified that (S n+1) is true. Hence, by PMI we know all (n ≥ 1) the statements (S n) are true. 2. Solution. We did this in class. 1 3. Solution. Let (S n): 2n3 + 3n2 + n is divisible by 6. (S 1) is just that 2 + 3 + 1 is divisible by 6, which is trivial. Assume that (S1), . . . , (Sn) are true. Prove that (Sn + 1) is true. But, 2(n + 1)3 + 3(n + 1)2 + (n + 1) = 2(n3 + 3n2 + 3n + 1) + 3(n2 + 2n + 1) + (n + 1) = (2n3 + 3n2 + n) + (6n2 + 6n + 2 + 6n + 3 + 1) = (2n3 + 3n2 + n) + 6(n2 + 2n + 1) The first term is divisible by 6 since (Sn) is true and the second term is a multiple of 6. Hence, the last quantity is divisible by 6. 4. Solution. (Fill in the details.) Check n = 1. Assume true for a particular n. That is, (1) is true. But, 1 · 2 + 2 · 3 + 3 · 4 + · · · n · (n + 1) + (n + 1)(n + 2) = = (n + 1)(n + 2) n(n + 1)(n + 2) + (n + 1)(n + 2) 3 (n + 1)(n + 2)(n + 3) n +1 = , 3 3 which shows that (S n+1) is true. 5. Solution. Recall f1 = 1, f2 = 1 and for n ≥ 2, fn = fn−2 + fn−1 . Check n = 1. Assume (2) is true. Then, n+1 X r=1 f2r−1 = n X f2r−1 + f2n+1 = (by the ind. hyp. )f2n + f2n+1 r=1 = f2(n+1) ( by the def. of Fib. ). 2