问题 解答题
设Sn等比数列{an}的前n项和,且a2=
1
9
S2=
4
9

(1)求数列{an}的通项;
(2)设bn=
n
an
,求数列{bn}的前n项和Sn
答案

(1)设首项为a1,公比为q,由a2=

1
9
,S2=
4
9

得:

a1q=
1
9
a1+a1q=
4
9

解得:

a1=
1
3
q=
1
3

∴an=

1
3n
;.

(2)∵bn=

n
an
=
n
1
3n
=n•3n

∴Sn=3+2×32+3×33+…+n•3n,①

∴3Sn=32+2×33+3×34+…+(n-1)•3n+n•3n+1,②

②-①得2Sn=n•3n+1-(3+32+33+…+3n)=n•3n+1-

3(1-3n)
1-3
=
(2n-1)×3n+1
2
+
3
2

∴Sn=

(2n-1)×3n+1
4
+
3
4

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