问题
解答题
设Sn等比数列{an}的前n项和,且a2=
(1)求数列{an}的通项; (2)设bn=
|
答案
(1)设首项为a1,公比为q,由a2=
,S2=1 9
,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•3n,n 1 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