问题 解答题
在数列{an}中,a1=1,当n≥2时,其前n项和Sn满足Sn2=an(Sn-
1
2
)

(1)求an
(2)令bn=
Sn
2n+1
,求数列{bn}的前项和Tn
答案

(1)当n≥2时,an=Sn-Sn-1

Sn2=(Sn-Sn-1)(Sn-

1
2
)=Sn2-
1
2
Sn-SnSn-1+
1
2
Sn-1

∴Sn-1-Sn=2SnSn-1

1
Sn
-
1
Sn-1
=2,

即数列{

1
Sn
}为等差数列,S1=a1=1,

1
Sn
=
1
S1
+(n-1)×2=2n-1,

Sn=

1
2n-1
,…(4分)

当n≥2时,an=sn-sn-1=

1
2n-1
-
1
2n-3
=
-2
(2n-1)(2n-3)

an

1,n=1
-2
(2n-1)(2n-3)
,n≥2
…(8分)

(2)bn=

Sn
2n+1
=
1
(2n-1)(2n+1)
=
1
2
(
1
2n-1
-
1
2n+1
)

Tn=

1
2
[(1-
1
3
)+(
1
3
-
1
5
)+…+(
1
2n-1
-
1
2n+1
)]=
1
2
(1-
1
2n+1
)=
n
2n+1

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