HW02
–
gilbert
–
(55035)
1
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find all
choices before answering.
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

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HW02
–
gilbert
–
(55035)
2
Compute the value of
5
correct
4a
n
b
n
2.
converges with limit
=
lim
4
when
n
→∞
3a
n
−
b
n
3.
converges with limit
=
5
7
n
lim
→∞
a
n
=
6,
n
lim
→∞
b
n
=
−
2.
4.
converges with limit
=
3
2
1.
limit
=
−
12
5
correct
5.
sequence does not converge
Explanation:
5
After division by n we see that
2.
limit
=
−
2
5 +
(
−
1)
n
3.
limit doesn’t exist
a
n
=
n
.
4.
limit
=
12
4 +
n
3
5
But
(
−
1)
n
,
3
5.
limit
=
5
−→
0
2
n
n
Explanation:
as n
→
∞
,
so a
n
→
5
4
as n
→
∞
.
Conse-
quently, the sequence converges and has
By properties of limits
lim 4a
n
b
n
limit
=
5
.
4
n
→
2
=
4
n
lim
→∞
a
n
n
lim
→∞
b
n
=
−
48
005
10.0 points
while
n
lim
→∞
(3a
n
−
b
n
)
Determine
if
the
sequence
{a
n
}
converges
when
a
n
=
5
n
2
(2n
−
1)!
,
=
3
n
lim
→∞
a
n
−
n
lim
→∞
b
n
=
20
6=
0.
(2n + 1)!
Thus, by properties of limits again,
and if it converges, find the limit.
1.
converges with limit
=
5
correct
4a
n
b
n
12
4
n
lim
→∞
3a
n
−
b
n
=
−
5
.
2.
converges with limit
=
2
5
004
10.0 points
5
2
Determine if the sequence {a
n
} converges,
3.
converges with limit
=
and if it does, find its limit when
4
4.
converges with limit
=
n
5
a
n
=
5n + (
−
1)
.
4n + 3
5.
does not converge
1.
converges with limit
=
1
Explanation: