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*).(1)求数列{an},{bn}的通项公式;(2)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*).(1)求数列{an},{bn}的通项公式;(2)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*).
(1)求数列{an},{bn}的通项公式;
(2)记cn=an•bn,求证:cn+1<cn
(3)求数列{cn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)∵等差数列{an}的公差大于0,且a3,a5是方程x2-14x+45=0的两根,
∴a3=5,a5=9,∴d=
a5-a3
5-3
=2
∴an=a5+2(n-5)=2n-1
∵Sn=
1-bn
2
,∴n≥2时,bn=Sn-Sn-1=
bn-1-bn
2
,∴
bn
bn-1
=class="stub"1
3

∵n=1时,b1=S1=
1-b1
2
,∴b1=class="stub"1
3

∴bn=class="stub"1
3
(class="stub"1
3
)n-1
=(class="stub"1
3
)
n

(2)证明:由(1)知cn=an•bn=class="stub"2n-1
3n

∴cn+1-cn=class="stub"2n+1
3n+1
-class="stub"2n-1
3n
=
4(1-n)
3n+1
≤0
∴cn+1<cn
(3)Tn=class="stub"1
3
+class="stub"3
32
+…+class="stub"2n-1
3n

class="stub"1
3
Tn=class="stub"1
32
+…+class="stub"2n-3
3n
+class="stub"2n-1
3n+1

两式相减可得:class="stub"2
3
Tn=class="stub"1
3
+class="stub"2
32
+…+class="stub"2
3n
-class="stub"2n-1
3n+1
=class="stub"3
2
-class="stub"3
2
•class="stub"n+1
3n

∴Tn=1-class="stub"n+1
3n

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