问题
解答题
已知{an}为等比数列,a1=1,前n项和为Sn,且
(1)求{an}和{bn}的通项公式; (2)设cn=an•bn,求{cn}的前n项和S′n. |
答案
(1)设等比数列的公比为q,则由
=28可知q≠1S6 S3
∵
=28,∴S6 S3
=1+q3=28,∴q=31-q6 1-q3
∵a1=1,∴an=3n-1
∵数列{bn}的前n项和为Tn,且点(n,Tn)均在抛物线y=
x2+1 2
x上1 2
∴Tn=
n2+1 2
n1 2
当n≥2时,bn=Tn-Tn-1= (
n2+1 2
n)-[1 2
(n-1)2+1 2
(n-1)]=n1 2
∵b1=T1=1
∴bn=n
(2)∵cn=an•bn=n•3n-1,∴S'n=1•30+2•31+3•32+…+n•3n-1,
∴3S'n=1•31+2•32+…+(n-1)•3n-1+n•3n,
两式相减,得-2S'n=1•30+1•31+1•32+…+1•3n-1-n•3n=
-n•3n=1-3n 1-3
-n•3n=3n-1 2
,(1-2n)3n-1 2
得 S'n=
.(2n-1)3n+1 4