已知等差数列{an}的公差大于0,且a3,a5是方程x2-14x+45=0的两根,数列{bn}的前n项的和为Sn,且Sn=1-bn2(n∈N*).(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)记cn

题目简介

已知等差数列{an}的公差大于0,且a3,a5是方程x2-14x+45=0的两根,数列{bn}的前n项的和为Sn,且Sn=1-bn2(n∈N*).(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)记cn

题目详情

已知等差数列{an}的公差大于0,且a3,a5是方程x2-14x+45=0的两根,数列{bn}的前n项的和为Sn,且Sn=
1-bn
2
(n∈N*)

(Ⅰ)求数列{an},{bn}的通项公式;
(Ⅱ)记cn=an•bn,求数列{cn}的前n项和Tn
题型:解答题难度:中档来源:太原模拟

答案

(Ⅰ)∵a3,a5是方程x2-14x+45=0的两根,且数列{an}的公差d>0,
∴a3=5,a5=9,公差d=
a5-a3
5-3
=2.

∴an=a5+(n-5)d=2n-1.(3分)
又当n=1时,有b1=S1=
1-b1
2

b1=class="stub"1
3

n≥2时,有bn=Sn-Sn-1=class="stub"1
2
(bn-1-bn),∴
bn
bn-1
=class="stub"1
3
(n≥2).

∴数列{bn}是首项b1=class="stub"1
3
,公比q=class="stub"1
3
等比数列,
bn=b1qn-1=class="stub"1
3n
.
(6分)

(Ⅱ)由(Ⅰ)知cn=anbn=class="stub"2n-1
3n
,则Tn=class="stub"1
31
+class="stub"3
32
+class="stub"5
33
++class="stub"2n-1
3n
(1)
class="stub"1
3
Tn
=class="stub"1
32
+class="stub"3
33
+class="stub"5
34
++class="stub"2n-3
3n
+class="stub"2n-1
3n+1
(2)(10分)
(1)-(2)得:class="stub"2
3
Tn=class="stub"1
3
+class="stub"2
32
+class="stub"2
33
++class="stub"2
3n
-class="stub"2n-1
3n+1
=class="stub"1
3
+2(class="stub"1
32
+class="stub"1
33
++class="stub"1
3n
)-class="stub"2n-1
3n+1

化简得:Tn=1-class="stub"n+1
3n
(12分)

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