Stitz-Zeager_College_Algebra_e-book

# At rst this problem seem misplaced but we can write

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Unformatted text preview: geometric. In an attempt to ﬁnd a pattern, we rewrite the second term with a denominator to make all the terms 4 See the footnotes on page 181 in Section 3.1 and page 330 of Section 6.1. 9.1 Sequences 557 2 appear as fractions. We have 5 , 2 , − 2 , − 2 , . . .. If we associate the negative ‘−’ of the last two 1 3 7 2 2 2 terms with the denominators we get 5 , 2 , −3 , −7 , . . .. This tells us that we can tentatively 1 sketch out the formula for the sequence as an = d2 where dn is the sequence of denominators. n Looking at the denominators 5, 1, −3, −7, . . ., we ﬁnd that they go from one term to the next by subtracting 4 which is the same as adding −4. This means we have an arithmetic sequence on our hands. Using Equation 9.1 with a = 5 and d = −4, we get the nth denominator by 2 the formula dn = 5 + (n − 1)(−4) = 9 − 4n for n ≥ 1. Our ﬁnal answer is an = 9−4n , n ≥ 1. 3. The sequence as given is neither arithmetic nor geometric, so we proceed as in the last problem to try to get patterns individually for the numerator and denominator. Letting cn and dn c denote the se...
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