题目内容
(1)计算
(1-
)(1-
)(1-
)…(1-
).
(2)若
(2n+
)=1,求
的值.
lim |
n→∞ |
1 |
22 |
1 |
32 |
1 |
42 |
1 |
4n2 |
(2)若
lim |
n→∞ |
an2-2n+1 |
bn+2 |
a |
b |
(1)(1-
)(1-
)(1-
)…(1-
)=
•
=
,
所以
(1-
)(1-
)(1-
)…(1-
)=
=
.
(2)2n+
=
,
且
(2n+
)=1,
所以
,
即
=-2.
1 |
22 |
1 |
32 |
1 |
42 |
1 |
4n2 |
1 |
2 |
2n+1 |
2n |
2n+1 |
4n |
所以
lim |
n→∞ |
1 |
22 |
1 |
32 |
1 |
42 |
1 |
4n2 |
lim |
n→∞ |
2n+1 |
4n |
1 |
2 |
(2)2n+
an2-2n+1 |
bn+2 |
(2b+a)n2+2n+1 |
bn+2 |
且
lim |
n→∞ |
an2-2n+1 |
bn+2 |
所以
|
即
a |
b |
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