问题
解答题
已知数列{an}的前n项和为Sn,a1=1,an=
(I)求a2,a3; (Ⅱ)设bn=
(Ⅲ)设cn=
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答案
(I)Sn+1=4an+2,当n=1时,a1+a2=4a1+2,a2=5;(1分)
当n=2时,a1+a2+a3=4a2+2,6+a3=22,a3=16;(2分)
(II)由an=
Sn+1-1 4
得,an+1=1 2
Sn+2-1 4
,an+1-an=1 2
an+21 4
an+1-an=1 2
an+2-1 4
an+1=1 2
(1 2
an+2-an+1),1 2
∴bn=
bn+1,1 2
=2∴数列{bn}是公比为2的等比数列.(4分)bn+1 bn
b1=
a2-a1=1 2
,3 2
∴bn=
•2n-1=3•2n-2(5分)3 2
(III)由(II)
3•2n-2=
an+1-an,1 2
=3 4
-an+1 2n+1
,令dn=an 2n
,d1=an 2n
=a1 2 1 2
∴数列{dn}是首项为
,公差为1 2
的等差数列.(7分)3 4
∴dn=
+(n-1)1 2
=3 4 3n-1 4
cn=
=22n+1 an•an+1
=1 dndn+1
(4 3
-1 dn
)1 dn+1
∴Tn=
(4 3
-1 d1
+1 d2
-1 d2
+…+1 d3
-1 dn
)=1 dn+1
(2-4 3
)=4 3n+2
(8分)8n 3n+2