Stitz-Zeager_College_Algebra_e-book

We rst rewrite 90 90 0 0 89 60 0 89 59 60 in essence

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Unformatted text preview: 3·2·1 = ++ + + 22 2 2 2 11 = + + 1 + 3 + 12 22 = 17 = (c) We proceed as before, replacing the index n, but not the variable x, with the values 1 through 5 and adding the resulting terms. 564 Sequences and the Binomial Theorem 5 n=1 (−1)n+1 (x − 1)n = n (−1)1+1 (−1)2+1 (−1)3+1 (x − 1)1 + (x − 1)2 + (x − 1)3 1 2 3 (−1)1+5 (−1)1+4 (x − 1)4 + (x − 1)5 + 4 5 (x − 1)2 (x − 1)3 (x − 1)4 (x − 1)5 = (x − 1) − + − + 2 3 4 5 2. The key to writing these sums with summation notation is to ﬁnd the pattern of the terms. To that end, we make good use of the techniques presented in Section 9.1. (a) The terms of the sum 1, 3, 5, etc., form an arithmetic sequence with ﬁrst term a = 1 and common diﬀerence d = 2. We get a formula for the nth term of the sequence using Equation 9.1 to get an = 1 + (n − 1)2 = 2n − 1, n ≥ 1. At this stage, we have the formula for the terms, namely 2n − 1, and the lower limit of the summation, n = 1. To ﬁnish the problem, we need to determine the upper limit of the summation. In other words, we need to dete...
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