题目内容
数列{an}满足a1=2,an+1=
(n∈N*).
(1)设bn=
,求数列{bn}的通项公式;
(2)设cn=
,数列{cn}的前n项和为Sn,求出Sn并由此证明:
≤Sn<
.
2n+1an | ||
(n+
|
(1)设bn=
2n |
an |
(2)设cn=
1 |
n(n+1)an+1 |
5 |
16 |
1 |
2 |
分析:(1)利用数列递推式,结合条件,可得bn+1-bn=n+
,利用叠加法,可求数列{bn}的通项公式;
(2)确定数列的通项,利用叠加法求和,利用数列的单调性,即可得到结论.
1 |
2 |
(2)确定数列的通项,利用叠加法求和,利用数列的单调性,即可得到结论.
解答:解:(1)∵an+1=
(n∈N*),
∴
-
=n+
∵bn=
∴bn+1-bn=n+
∴bn=b1+(b2-b1)+…+(bn-bn-1)=b1+
∵bn=
,a1=2,
∴b1=1
∴bn=
;
(2)由(1)知,an=
,∴an+1=
,
∴cn=
=
[
+
-
]
∴Sn=
×
+
[
-
]=
[1-(
)n+1×
]
∵(
)n+1×
=(
)n+1×(1+
)得到递减,
∴(
)n+1×
≤(
)1+1×
=
∴
≤
[1-(
)n+1×
]<
,即
≤Sn<
.
2n+1an | ||
(n+
|
∴
2n+1 |
an+1 |
2n |
an |
1 |
2 |
∵bn=
2n |
an |
∴bn+1-bn=n+
1 |
2 |
∴bn=b1+(b2-b1)+…+(bn-bn-1)=b1+
n2-1 |
2 |
∵bn=
2n |
an |
∴b1=1
∴bn=
n2+1 |
2 |
(2)由(1)知,an=
2n+1 |
n2+1 |
2n+2 |
(n+1)2+1 |
∴cn=
1 |
n(n+1)an+1 |
1 |
2 |
1 |
2n+1 |
1 |
n×2n |
1 |
(n+1)×2n+1 |
∴Sn=
1 |
2 |
| ||||
1-
|
1 |
2 |
1 |
2 |
1 |
(n+1)×2n+1 |
1 |
2 |
1 |
2 |
n+2 |
n+1 |
∵(
1 |
2 |
n+2 |
n+1 |
1 |
2 |
1 |
n+1 |
∴(
1 |
2 |
n+2 |
n+1 |
1 |
2 |
1+2 |
1+1 |
3 |
8 |
∴
5 |
16 |
1 |
2 |
1 |
2 |
n+2 |
n+1 |
1 |
2 |
5 |
16 |
1 |
2 |
点评:本题考查数列的通项与求和,考查叠加法的运用,考查学生分析解决问题的能力,属于中档题.
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