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)求an; (2)令bn=
|
答案
(1)当n≥2时,an=Sn-Sn-1,
∴Sn2=(Sn-Sn-1)(Sn-
)=Sn2-1 2
Sn-SnSn-1+1 2
Sn-1,1 2
∴Sn-1-Sn=2SnSn-1,
∴
-1 Sn
=2,1 Sn-1
即数列{
}为等差数列,S1=a1=1,1 Sn
∴
=1 Sn
+(n-1)×2=2n-1,1 S1
∴Sn=
,…(4分)1 2n-1
当n≥2时,an=sn-sn-1=
-1 2n-1
=1 2n-3 -2 (2n-1)(2n-3)
∴an=
…(8分)1,n=1
,n≥2-2 (2n-1)(2n-3)
(2)bn=
=Sn 2n+1
=1 (2n-1)(2n+1)
(1 2
-1 2n-1
),1 2n+1
∴Tn=
[(1-1 2
)+(1 3
-1 3
)+…+(1 5
-1 2n-1
)]=1 2n+1
(1-1 2
)=1 2n+1 n 2n+1