题目内容
(2012•杭州一模)在数列{an}中,a1=1,当n≥2时,其前n项和Sn满足Sn2=an(Sn-
).
(1)求an;
(2)令bn=
,求数列{bn}的前项和Tn.
1 |
2 |
(1)求an;
(2)令bn=
Sn |
2n+1 |
分析:(1)当n≥2时,由an=Sn-Sn-1,代入已知整理可得Sn-1-Sn=2SnSn-1,即
-
=2,结合等差数列的通项公式可求Sn,进而可求当n≥2时an,在对n=1时求a1,从而可求an
(2)由于bn=
=
=
(
-
),考虑利用裂项求和即可
1 |
Sn |
1 |
Sn-1 |
(2)由于bn=
Sn |
2n+1 |
1 |
(2n-1)(2n+1) |
1 |
2 |
1 |
2n-1 |
1 |
2n+1 |
解答:解:(1)当n≥2时,an=Sn-Sn-1,
∴Sn2=(Sn-Sn-1)(Sn-
)=Sn2-
Sn-SnSn-1+
Sn-1,
∴Sn-1-Sn=2SnSn-1,
∴
-
=2,
即数列{
}为等差数列,S1=a1=1,
∴
=
+(n-1)×2=2n-1,
∴Sn=
,…(4分)
当n≥2时,an=sn-sn-1=
-
=
∴an=
…(8分)
(2)bn=
=
=
(
-
),
∴Tn=
[(1-
)+(
-
)+…+(
-
)]=
(1-
)=
∴Sn2=(Sn-Sn-1)(Sn-
1 |
2 |
1 |
2 |
1 |
2 |
∴Sn-1-Sn=2SnSn-1,
∴
1 |
Sn |
1 |
Sn-1 |
即数列{
1 |
Sn |
∴
1 |
Sn |
1 |
S1 |
∴Sn=
1 |
2n-1 |
当n≥2时,an=sn-sn-1=
1 |
2n-1 |
1 |
2n-3 |
-2 |
(2n-1)(2n-3) |
∴an=
|
(2)bn=
Sn |
2n+1 |
1 |
(2n-1)(2n+1) |
1 |
2 |
1 |
2n-1 |
1 |
2n+1 |
∴Tn=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
2n-1 |
1 |
2n+1 |
1 |
2 |
1 |
2n+1 |
n |
2n+1 |
点评:本题主要考查了利用递推公式an=
求解数列的通项公式,要注意对n=1的检验是做题中容易漏掉的知识点,还考查了裂项求和方法的应用.
|

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