VIEWS: 4 PAGES: 3 POSTED ON: 8/7/2012 Public Domain
Advanced Mathematical Concepts Chapter 12 Lesson 12-9 Example 1 n(3n - 1) Prove Sn: 1 + 4 + 7 + … + (3n - 2) = . 2 1. First, verify that Sn is valid for n = 1. 1[3(1) - 1] Since S1 = 1 and 2 = 1, the formula is valid for n = 1. 2. Then, assume that Sn is valid for n = k. k(3k - 1) Sk 1 + 4 + 7 + ... + (3k - 2) = 2 Replace n with k. Next, prove that it is also valid for n = k + 1. k(3k - 1) Sk + 1 1 + 4 + 7 + ... + (3k - 2) + (3k + 1) = 2 + (3k + 1) 3(k + 1) - 2 = 3k + 1 k(3k - 1) 2(3k + 1) = + 2 2 2 3k - k + 6k + 2 = 2 3k2 + 5k + 2 = 2 (k + 1)(3k + 2) = 2 If k + 1 is substituted into the original formula, the same result is obtained. Thus, if the formula is valid for n = k, it is also valid for n = k + 1. Since Sn is valid for n = 1, it is also valid for n = 2, n = 3, and so on. Advanced Mathematical Concepts Chapter 12 Example 2 Prove that 8n – 1 is divisible by 7 for all positive integers n. Using the definition of divisibility, we can state the conjecture as follows. Sn 8n - 1 = 7r for some integer r. 1. First verify that this is valid for n = 1. S1 81 - 1 = 7 Since 7 is divisible by 7, this is valid for n = 1. 2. Then assume that Sn is valid for n = k and use this assumption to prove that it is also valid for n = k + 1. Sk 8k - 1 = 7r for some integer r. Assume sk is true. Sk + 1 8k + 1 - 1 = 7t for some integer t. Show that Sk + 1 must follow. For this proof, rewrite the left-hand side of Sk so that it matches the left-hand side of Sk + 1. 8k - 1 = 7r Sk 8(8k - 1) = 8(7r) Multiply each side by 8. 8k + 1 - 8 = 56r Simplify. 8k + 1 - 1 = 56r + 7 Add 7 to each side. 8k + 1 - 1 = 7(8r + 1) Factor. Let t = 8r + 1, an integer. Then 8k + 1 - 1 = 7t. We have shown that if Sk is valid, then Sk + 1 is also valid. Since Sn is valid for n = 1, it is also valid for n = 2, n = 3, and so on. Hence, 8n – 1 is divisible by 7 for all positive integers n. Advanced Mathematical Concepts Chapter 12 Example 3 Prove Sn: 1 + 3 + 5 + 7 + … + (2n – 1) = n2. 1. First verify that Sn is valid for the first possible case, n = 1. If n = 1, then S1 is 1. The sum is 1 and 12 = 1, so the formula is valid for the first case. 2. Then assume that the formula is valid for n = k and prove that it is also valid for n = k + 1. It must be proven that if Sk: 1 + 3 + 5 + 7 + … = k2, then Sk + 1: 1 + 3 + 5 + 7 + … = (k + 1)2 is also valid. The kth term of the series is 2k – 1. Substitute k + 1 for k. Then 2(k + 1) – 1 = 2k + 1. You can write the formula for Sk+1 by adding the next term, 2k + 1, to each side. Sk 1 + 3 + 5 + 7 + ... + (2k - 1) = k2 Sk + 1 1 + 3 + 5 + 7 + ... + (2k - 1) + (2k + 1) = k2 + (2k + 1) = (k + 1)2 Since the formula for the sum for n = k + 1 produces the same result as the direct computation of the sum of the series, the formula is valid for n = k + 1 if it is valid for n = k. Thus, it can be concluded that since the formula is valid for n = 1, it is also valid for n = 2. Since it is valid for n = 2, it is valid for n = 3, and so on, indefinitely. Therefore, the formula is valid for any positive integer n.