设数列{an}为等差数列,且a5=14,a7=20,数列{bn}的前n项和为Sn=1-(13)n(n∈N*),(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)若cn=an•bn,n=1,2,3,…,求

题目简介

设数列{an}为等差数列,且a5=14,a7=20,数列{bn}的前n项和为Sn=1-(13)n(n∈N*),(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)若cn=an•bn,n=1,2,3,…,求

题目详情

设数列{an} 为等差数列,且a5=14,a7=20,数列{bn} 的前n项和为Sn=1-(
1
3
)
n
(n∈N*),
(Ⅰ)求数列{an},{bn}的通项公式;
(Ⅱ)若cn=an•bn,n=1,2,3,…,求数列{cn}的前n项和Tn
题型:解答题难度:中档来源:邯郸二模

答案

(Ⅰ)∵数列{an}为等差数列,且a5=14,a7=20,
∴公差d=class="stub"1
2
(a7-a5)
=3,
∵a5=a1+4×3=14,
∴a1=2.
∴an=2+(n-1)×3=3n-1.
∵数列{bn}的前n项和为Sn=1-(class="stub"1
3
)
n
(n∈N*),
b1=S1=1-class="stub"1
3
=class="stub"2
3

bn=Sn-Sn-1=[1-(class="stub"1
3
)
n
]-[1-(class="stub"1
3
)
n-1
]=class="stub"2
3n

当n=1时,class="stub"2
3n
=class="stub"2
3
=a1

bn=class="stub"2
3 n

(Ⅱ)由an=3n-1,bn=class="stub"2
3 n

得cn=an•bn=2(3n-1)•class="stub"1
3n

Tn=2[2•class="stub"1
3
+5•class="stub"1
3 2
+8•class="stub"1
3 3
+…+(3n-1)•class="stub"1
3 n
]

class="stub"1
3
Tn=2[2•class="stub"1
3 2
+5•class="stub"1
33
+…+
(3n-4)•class="stub"1
3 n
+(3n-1)•class="stub"1
3 n+1
]

两式相减,得class="stub"2
3
Tn=2[3•class="stub"1
3
+3•class="stub"1
3 2
+3•class="stub"1
3 3
+…+
+3•class="stub"1
3 n
-class="stub"1
3
-(3n-1)•class="stub"1
3 n+1
]

Tn=class="stub"7
2
-class="stub"7
2
•class="stub"1
3 n
-class="stub"n
3 n-1

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