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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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