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Stitz-Zeager_College_Algebra_e-book

# 1 in the process k1 j 0 k kj j 1 ab j k11 k akj 1

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Unformatted text preview: ce given in (1). 3 We know from Example 9.1.2 that it is geometric with common ratio r = − 2 . The ﬁrst term is n−1 3 for n ≥ 1. After a touch of simplifying, a = 1 so by Equation 9.1 we get an = arn−1 = 1 − 2 2 2 n−1 we get an = (−3)n for n ≥ 1. Note that we can easily check our answer by substituting in values 2 of n and seeing that the formula generates the sequence given in (1). We leave this to the reader. Our next example gives us more practice ﬁnding patterns. Example 9.1.3. Find an explicit formula for the nth term of the following sequences. 1. 0.9, 0.09, 0.009, 0.0009, . . . 2. 2 22 , 2, − , − , . . . 5 37 24 8 3. 1, − , , − , . . . 7 13 19 Solution. 1. Although this sequence may seem strange, the reader can verify it is actually a geometric 1 9 9 1 n−1 sequence with common ratio r = 0.1 = 10 . With a = 0.9 = 10 , we get an = 10 10 for 9 n ≥ 0. Simplifying, we get an = 10n , n ≥ 1. There is more to this sequence than meets the eye and we shall return to this example in the next section. 2. As the reader can verify, this sequence is neither arithmetic nor...
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