Stitz-Zeager_College_Algebra_e-book

2 d k 2 k1 k 1 2 prove n 3n for all n 7 solution 1

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Unformatted text preview: + 1 = 2 − a1 which simplifies to a2 = 2 − a1 = 2 − 7 = −5. When n = 2, the equation becomes a2 + 1 = 2 − a2 so we get a3 = 2 − a2 = 2 − (−5) = 7. Finally, when n = 3, we get a3 + 1 = 2 − a3 so a4 = 2 − a3 = 2 − 7 = −5. 6. As with the problem above, we are given a place to start with f0 = 1 and given a formula to build other terms of the sequence. Substituting n = 1 into the equation fn = n · fn−1 , we get f1 = 1 · f0 = 1 · 1 = 1. Advancing to n = 2, we get f2 = 2 · f1 = 2 · 1 = 2. Finally, f3 = 3 · f2 = 3 · 2 = 6. Some remarks about Example 9.1.1 are in order. We first note that since sequences are functions, we can graph them in the same way we graph functions. For example, if we wish to graph the sequence {bk }∞ from Example 9.1.1, we graph the equation y = b(k ) for the values k ≥ 0. That k=0 is, we plot the points (k, b(k )) for the values of k in the domain, k = 0, 1, 2, . . .. The resulting collection of points is the graph of the sequence. Note that...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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