HW02 – gilbert – (55035)
1
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001
10.0 points
Find a formula for the general term a
n
of the sequence
assuming that the pattern of the first few
terms continues.
₃
2
₃
n
2.
a
n
=
−
3
₃
3
₃
n
3.
a
n
=
−
2
ë
o
{a
n
}
∞
n=1
=
2,
6,
10,
14,
. . .
,
assuming that the pattern of the first
terms continues.
1. a
n
=
n + 3
2. a
n
=
4n − 2 correct
3. a
n
=
5n − 3
4. a
n
=
3n − 1
5. a
n
=
n + 4
Explanation:
By inspection, consecutive terms
a
n−1
a
n
in the sequence
few
and
3.
a
n
=
₃
−
2
3
₃
n−1
4.
a
n
=
₃
−
3
2
₃
n−1
correct
₃
3
₃
n−1
6.
a
n
=
−
4
₃
3
₃
n
7.
a
n
=
−
4
Explanation:
By inspection, consecutive terms a
n−1
and a
n
in the sequence
{a
n
}
∞
n=1
=
ë
1,
−
2
3
,
4
9
,
−
27
8
,
. . .
o
ë
o
{a
n
}
∞
n=1
=
2,
6,
10,
14,
. . .
have the property that
a
n
− a
n−1
=
d
=
4 .
Thus
a
n
=
a
n−1
+ d
=
a
n−2
+ 2d
=
. . .
=
a
1
+ (n − 1)d
=
2 + 4(n − 1) .
Consequently,
a
n
=
4n − 2
.
keywords:
002
10.0 points
Find
a
formula
for
the
general
term a
n
of
the sequence
{a
n
}
n
∞
=1
=
ë
1,
−
3
2
,
9
4
,
−
27
8
,
. . .
o
,
have the property that
₃
2
₃
a
n
=
ra
n−1
=
−
3
a
n−1
.
Thus
a
n
=
ra
n−1
= r
2
a
n−2
=
. . .
=
r
n−1
a
1
=
₃
−3
2
₃
n−1
a
1
.
Consequently,
a
n
=
₃
−3
2
₃
n−1
since a
1
= 1.
keywords:
sequence, common ratio
003 10.0 points