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Pre-Calc Homework Solutions 453

Pre-Calc Homework Solutions 453 - Appendix A3(b Step 1 The...

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(b) Step 1: The formula holds for n 5 1, because 1 k 5 1 ( a k 2 b k ) 5 1 k 5 1 a k 2 1 k 5 1 b k 5 a 1 2 b 1 . Step 2: Suppose i k 5 1 ( a k 2 b k ) 5 i k 5 1 a k 2 i k 5 1 b k . Then i 1 1 k 5 1 ( a k 2 b k ) 5 3 i k 5 1 ( a k 2 b k ) 4 1 ( a i 1 1 2 b i 1 1 ) 5 3 i k 5 1 a k 4 2 3 i k 5 1 b k 4 1 a i 1 1 2 b i 1 1 5 i 1 1 k 5 1 a k 2 i 1 1 k 5 1 b k . The mathematical induction principle now guarantees the original formula for every positive integer n . (c) Step 1: The formula holds for n 5 1, because 1 k 5 1 ca k 5 c ? 1 k 5 1 a k 5 ca 1 . Step 2: Suppose i k 5 1 ca k 5 c ? i k 5 1 a k . Then i 1 1 k 5 1 ca k 1 1 5 3 i k 5 1 ca k 4 1 ca k 1 1 5 3 c ? i k 5 1 a k 4 1 ca k 1 1 5 c 31 i k 5 1 a k 2 1 a k 1 1 4 5 c ? i 1 1 k 5 1 a k 1 1 . The mathematical induction principle how guarantees the original formula for every positive integer n . (d) Step 1: The formula n k 5 1 c 5 n ?
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