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Mathematical Induction (10 points) To prove a theorem holds true for every natural number, n, you can use induction. 1. Prove it is true for n = 1 (or some specific number.) This is called the “base” case. 2. Assuming the theorem is true for n = k (this is called the “induction hypothesis”.) 3. Now do the work to show the next case follows. That is, if the theorem is true for n = k, then it is also true for n = k+1. Here is a website that I found that might help: http://www.themathpage.com/aprecalc/mathematical-induction.htm n(n + 1) They have a nice example proving that 13 + 2 3 + ... + n 3 = and another proving that 2 the sum of the first n odd numbers is equal to the square of n. 2 Here are 2 problems for you to do: 1. (4 points) Prove that: 1 ⋅ 3 + 2 ⋅ 4 + 3 ⋅ 5 + ... + n(n + 2) = n(n + 1)(2n + 7) 6 2. (6 points) Prove by using mathematical induction: For any positive integer n, 5 6 n+4 n(3n + 7) + +...+ = 1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4 n(n + 1)(n + 2) 2(n + 1)(n + 2) Here are some nice formulas: Sigma Notation = Closed Form Expanded 1 + 1 + 1 + ... + 1 (n times) 1 + 2 + 3 + ... + n 1 + 4 + 9 + ... + n2 1 + 8 + 27 + ... + n3 1 + 16 + 81 + ... + n4 1 + 32 + 243 + ... + n5