Pre-Calc Homework Solutions 452

# Pre-Calc Homework - 452 Appendix A2 9 Step 1 The formula holds for n 3 6 27 4 5 6 7 343 6 because 12 1(1 1/2(1 3 1 6 Experiment n n n3 1 1 1 2 2 8

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6. Experiment: Step 1: The inequality holds for n 5 6, because 720 . 216. Step 2: Suppose k ! . k 3 . Then ( k 1 1)! . ( k 1 1) k 3 . For k \$ 4, k . 1 1 } 2 k } 1 } k 1 2 } (since } 2 k } , 2 and } k 1 2 } , 1), and so k 3 . k 2 1 2 k 1 1 5 ( k 1 1) 2 . So ( k 1 1) k 3 . ( k 1 1) 3 , and thus by the transitivity of . ,( k 1 1)! . ( k 1 1) 3 . The mathematical induction principle now guarantees the original inequality for all n \$ 6. 7. Experiment: Step 1: The inequality holds for n 5 5, because 32 . 25. Step 2: Suppose 2 k . k 2 . For k \$ 3, k . 2 1 } 1 k } 1 since } 1 k } , 1 2 , and so k 2 . 2 k 1 1. Then by the transitivity of . ,2 k . 2 k + 1. And thus 2 k 1 1 5 2 k 1 2 k . k 2 1 2 k 1 1 5 ( k 1 1) 2 . The mathematical induction principle now guarantees the original inequality for all n \$ 5. 8. Step 1: The inequality holds for n 52 3, because 2 2 3 5 } 1 8 } . Step 2: Suppose 2 k \$ } 1 8 } . Then 2 k 1 1 5 2 ? 2 k \$ } 2 8 } 5 } 1 4 } , and by the transitivity of \$ and the fact that } 1 4 } \$ } 1 8 } k 1 1 \$ } 1 8 } . The mathematical induction principle now guarantees the original inequality for n \$2 3.
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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