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Stitz-Zeager_College_Algebra_e-book

# In the scenario of example 1015 the period of the

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Unformatted text preview: t P (k + 1) is true. Since (z )k+1 = (z )k z , we can use the induction hypothesis and 576 Sequences and the Binomial Theorem write (z )k = z k . Hence, (z )k+1 = (z )k z = z k z . We now use the product rule for conjugates4 to write z k z = z k z = z k+1 . This establishes (z )k+1 = z k+1 , so that P (k + 1) is true. Hence, by the Principle of Mathematical Induction, (z )n = z n for all n ≥ 1. 3. The ﬁrst wrinkle we encounter in this problem is that we are asked to prove this formula for n > 5 instead of n ≥ 1. Since n is a natural number, this means our base step occurs at n = 6. We can still use the PMI in this case, but our conclusion will be that the formula is valid for all n ≥ 6. We let P (n) be the inequality 3n > 100n, and check that P (6) is true. Comparing 36 = 729 and 100(6) = 600, we see 36 > 100(6) as required. Next, we assume that P (k ) is true, that is we assume 3k > 100k . We need to show that P (k + 1) is true, that is, we need to show 3k+1 > 100(k + 1). Since...
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