a2,a5是方程x2-12x+27=0的两根,数列{an}是公差为正的等差数列,数列{bn}的前n项和为Tn,且Tn=1-12bn(n∈N*).(1)求数列{an},{bn}的通项公式;(2)记cn=

题目简介

a2,a5是方程x2-12x+27=0的两根,数列{an}是公差为正的等差数列,数列{bn}的前n项和为Tn,且Tn=1-12bn(n∈N*).(1)求数列{an},{bn}的通项公式;(2)记cn=

题目详情

a2,a5是方程x2-12x+27=0的两根,数列{an}是公差为正的等差数列,数列{bn}的前n项和为Tn,且Tn=1-
1
2
bn(n∈N*).
(1)求数列{an},{bn}的通项公式;  
(2)记cn=anbn,求数列{cn}的前n项和Sn
题型:解答题难度:中档来源:广州一模

答案

(1)由a2+a5=12,a2•a5=27,且d>0,得a2=3,a5=9,∴d=
a5-a2
3
=2,a1=1,∴an=2n-1,
在Tn=1-class="stub"1
2
bn,令n=1,得b1=class="stub"2
3
,当n≥2时,Tn=1-class="stub"1
2
 bn 中,令 n=1得 b1=class="stub"2
3
,当n≥2时,
Tn=1-class="stub"1
2
bn,Tn-1=1-class="stub"1
2
bn-1
,两式相减得 bn= class="stub"1
2
bn-1-class="stub"1
2
bn
bn
bn-1
=class="stub"1
3
 (n≥2),
bn=class="stub"2
3
(class="stub"1
3
)
n-1
 
=class="stub"2
3n
  (n∈N+).
(2)cn= (2n-1)class="stub"2
3n
=class="stub"4n-2
3n
,∴Sn=2(class="stub"1
3
+class="stub"3
32
 +class="stub"5
33
+…+class="stub"2n-1
3n
),
class="stub"1
3
Sn=2(class="stub"1
32
+class="stub"3
33
+…+  class="stub"2n-3
3n
+ class="stub"2n-1
3n+1
 ),
 两式相减可解得  Sn=2-class="stub"2n+2
3n

更多内容推荐