Stitz-Zeager_College_Algebra_e-book

4 the binomial theorem 585 we forgo the simplication

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Unformatted text preview: atios 5 25 125 5 a3 5 5 a2 a4 =9= , = 27 = = 81 = and 1 5 25 a1 3 a2 3 a3 3 3 9 27 suggest that the sequence is geometric. To prove it, we must show that an+1 an an+1 an = r for all n. 5(n+1)−1 n+1 5n 3n 5 = 3n−1 = n+1 · n−1 = 3 5 3 5 3n This sequence is geometric with common ratio r = 5 . 3 2. Again, we have Example 9.1.1 to thank for providing the first four terms of this sequence: 1 8 1, − 1 , 1 and − 7 . We find b1 − b0 = − 4 and b2 − b1 = 15 . Hence, the sequence is not 35 3 1 arithmetic. To see if it is geometric, we compute b1 = − 3 and b2 = − 3 . Since there is no b0 b1 5 ‘common ratio,’ we conclude the sequence is not geometric, either. 3. As we saw in Example 9.1.1, the sequence {2n − 1}∞ generates the odd numbers: 1, 3, 5, 7, . . .. n=1 Computing the first few differences, we find a2 − a1 = 2, a3 − a2 = 2, and a4 − a3 = 2. This suggests that the sequence is arithmetic. To verify this, we find an+1 − an = (2(n + 1) − 1) − (2n − 1) = 2n + 2 − 1 − 2n + 1 = 2 This establishes that...
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