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Stitz-Zeager_College_Algebra_e-book

# Now the entries on the main diagonal of a11 are the

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Unformatted text preview: Now suppose we wanted to know a117 , that is, the 117th term in the sequence. 8 While the pattern of the sequence is apparent, it would beneﬁt us greatly to have an explicit formula for an . Unfortunately, there is no general algorithm that will produce a formula for every sequence, so any formulas we do develop will come from that greatest of teachers, experience. In other words, it is time for an example. Example 9.1.1. Write the ﬁrst four terms of the following sequences. 1. an = 2. bk = 5n−1 ,n≥1 3n (−1)k 2k + 1 ,k≥0 3. {2n − 1}∞ n=1 4. 1 + (−1)j j ∞ j =2 5. a1 = 7, an+1 = 2 − an , n ≥ 1 6. f0 = 1, fn = n · fn−1 , n ≥ 1 Solution. 1. Since we are given n ≥ 1, the ﬁrst four terms of the sequence are a1 , a2 , a3 and a4 . Since the notation a1 means the same thing as a(1), we obtain our ﬁrst term by replacing every 1−1 1 occurrence of n in the formula for an with n = 1 to get a1 = 5 31 = 3 . Proceeding similarly, 2−1 3−1 4−1 5 we get a2 = 5 32 = 9 , a3 = 5 33 = 25 and a4 = 5 34 = 125 . 27 81 2. For this sequence we have k ≥ 0, so the ﬁrst four terms are b0 , b1 , b2 and b3 . Pr...
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