已知{an}为等比数列,a1=1,前n项和为Sn,且S6S3=28,数列{bn}的前n项和为Tn,且点(n,Tn)均在抛物线y=12x2+12x上.(1)求{an}和{bn}的通项公式;(2)设cn=

题目简介

已知{an}为等比数列,a1=1,前n项和为Sn,且S6S3=28,数列{bn}的前n项和为Tn,且点(n,Tn)均在抛物线y=12x2+12x上.(1)求{an}和{bn}的通项公式;(2)设cn=

题目详情

已知{an}为等比数列,a1=1,前n项和为Sn,且
S6
S3
=28
,数列{bn}的前n项和为Tn,且点(n,Tn)均在抛物线y=
1
2
x2+
1
2
x
上.
(1)求{an}和{bn}的通项公式;
(2)设cn=an•bn,求{cn}的前n项和S′n
题型:解答题难度:中档来源:蓝山县模拟

答案

(1)设等比数列的公比为q,则由
S6
S3
=28
可知q≠1
S6
S3
=28
,∴
1-q6
1-q3
=1+q3=28
,∴q=3
∵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=class="stub"1
2
x2+class="stub"1
2
x

Tn=class="stub"1
2
n2+class="stub"1
2
n

当n≥2时,bn=Tn-Tn-1= (class="stub"1
2
n2+class="stub"1
2
n)-[class="stub"1
2
(n-1)2+class="stub"1
2
(n-1)]
=n
∵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=
1-3n
1-3
-n•3n=
3n-1
2
-n•3n=
(1-2n)3n-1
2

得  S'n=
(2n-1)3n+1
4

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