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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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