题目内容
已知在数列{an}中,a1=1,当n≥2时,其前n项和Sn满足Sn2=an(Sn-1 |
2 |
(Ⅰ) 求Sn的表达式;
(Ⅱ) 设bn=
Sn |
2n+1 |
分析:(Ⅰ)当n≥2时,把an=Sn-Sn-1代入Sn2=an(Sn-
)即可得到2SnSn-1+Sn-Sn-1=0,然后化简得
-
=2,于是可以得到Sn的表达式,
(Ⅱ)把Sn=
代入bn=
中可得bn=
=
(
-
),然后进行裂项相消进行求和.
1 |
2 |
1 |
Sn |
1 |
Sn-1 |
(Ⅱ)把Sn=
1 |
2n-1 |
Sn |
2n+1 |
1 |
(2n-1)(2n+1) |
1 |
2 |
1 |
2n-1 |
1 |
2n+1 |
解答:解:(Ⅰ)当n≥2时,an=Sn-Sn-1代入得:2SnSn-1+Sn-Sn-1=0?
-
=2,
∴
=2n-1?Sn=
(6分)
(Ⅱ)bn=
=
=
(
-
)
∴Tn=
[(
-
)+(
-
)+…+(
-
)]=
.(13分)
1 |
Sn |
1 |
Sn-1 |
∴
1 |
Sn |
1 |
2n-1 |
(Ⅱ)bn=
Sn |
2n+1 |
1 |
(2n-1)(2n+1) |
1 |
2 |
1 |
2n-1 |
1 |
2n+1 |
∴Tn=
1 |
2 |
1 |
1 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
2n-1 |
1 |
2n+1 |
n |
2n+1 |
点评:本题主要考查数列的求和和求数列递推式的知识点,利用裂项相消法求数列的和是解答本题第二问的关键,本题难度一般.
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