题目内容
已知数列{an}满足:a1=1且an+1=3+4an |
12-4an |
(1)若数列{bn}满足:bn=
1 | ||
an-
|
(2)求数列{anbn}的前n项和Sn;
(3)数列{an-bn}是否存在最大项,如果存在求出,若不存在说明理由.
分析:(1)由
=
=
=
=
.知数列{bn-1}是等比数列.
(2)由bn=
,得anbn=1+
bn,知anbn=1+
[1+(
)n-1]=
+
(
)n-1,由此知Sn=
[
+
(
)n-1]=
n+
=
n+
(
)n-
.
(3)由bn=
,得an=
+
,∴an-bn=
+
-bn=
-bn+
、又由(2)知,bn=1+(
)n-1,数列{bn}单调递增,所以数列an-bn为单调递减数列,由此知数列an-bn中存在最大项且为该数列中的首项,其值为-1.
bn+1-1 |
bn-1 |
| ||||
|
| ||||||
|
| ||
|
5 |
3 |
(2)由bn=
1 | ||
an-
|
1 |
2 |
1 |
2 |
5 |
3 |
3 |
2 |
1 |
2 |
5 |
3 |
n |
k=1 |
3 |
2 |
1 |
2 |
5 |
3 |
3 |
2 |
| ||||
|
3 |
2 |
3 |
4 |
5 |
3 |
3 |
4 |
(3)由bn=
1 | ||
an-
|
1 |
bn |
1 |
2 |
1 |
bn |
1 |
2 |
1 |
bn |
1 |
2 |
5 |
3 |
解答:解:(1)∵
=
=
=
=
.
∴数列{bn-1}是等比数列,首项为b1-1=
-1=1,公比为
.
(2)由bn=
,得anbn=1+
bn.
由(1)得bn-1=(
)n-1, ∴bn=1+(
)n-1,
∴anbn=1+
[1+(
)n-1]=
+
(
)n-1,
∴Sn=
[
+
(
)n-1]=
n+
=
n+
(
)n-
.
(3)由bn=
,得an=
+
,
∴an-bn=
+
-bn=
-bn+
,
又由(2)知,bn=1+(
)n-1,
∴数列{bn}单调递增,∴{
}与-bn均为递减数列、∴数列{an-bn}为单调递减数列,
∴当n=1时,a1-b1=1-2=-1最大,即数列{an-bn}中存在最大项且为该数列中的首项,其值为-1、
bn+1-1 |
bn-1 |
| ||||
|
| ||||||
|
| ||
|
5 |
3 |
∴数列{bn-1}是等比数列,首项为b1-1=
1 | ||
a1-
|
5 |
3 |
(2)由bn=
1 | ||
an-
|
1 |
2 |
由(1)得bn-1=(
5 |
3 |
5 |
3 |
∴anbn=1+
1 |
2 |
5 |
3 |
3 |
2 |
1 |
2 |
5 |
3 |
∴Sn=
n |
k=1 |
3 |
2 |
1 |
2 |
5 |
3 |
3 |
2 |
| ||||
|
3 |
2 |
3 |
4 |
5 |
3 |
3 |
4 |
(3)由bn=
1 | ||
an-
|
1 |
bn |
1 |
2 |
∴an-bn=
1 |
bn |
1 |
2 |
1 |
bn |
1 |
2 |
又由(2)知,bn=1+(
5 |
3 |
∴数列{bn}单调递增,∴{
1 |
bn |
∴当n=1时,a1-b1=1-2=-1最大,即数列{an-bn}中存在最大项且为该数列中的首项,其值为-1、
点评:本题考查数列的性质和应用,解题时要认真审题,仔细解答.
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