等差数列{an}中,a3=7,a1+a2+a3=12,令bn=anan+1,数列{1bn}的前n项和为Tn.(1)求数列{an}的通项公式.(2)求证:Tn<13-数学

题目简介

等差数列{an}中,a3=7,a1+a2+a3=12,令bn=anan+1,数列{1bn}的前n项和为Tn.(1)求数列{an}的通项公式.(2)求证:Tn<13-数学

题目详情

等差数列{an}中,a3=7,a1+a2+a3=12,令bn=anan+1,数列{
1
bn
}
的前n项和为Tn
(1)求数列{an}的通项公式.
(2)求证:Tn
1
3
题型:解答题难度:中档来源:不详

答案

(1)设数列{an}的公差为d,
∵a3=7,a1+a2+a3=12
a1+2d=7
3a1+3d=12

解得
a1=1
d=3

∴数列{an}的通项公式为:an=3n-2(n∈N*)
(2)∵bn=anan-1,
∴bn=(3n-2)(3n+1)
class="stub"1
bn
=class="stub"1
3
(class="stub"1
3n-2
-class="stub"1
3n+1
)

∴数列{class="stub"1
bn
}
的前n项和
Tn=class="stub"1
3
[1-class="stub"1
4
+class="stub"1
4
-class="stub"1
7
+class="stub"1
7
-class="stub"1
11
++class="stub"1
3n-5
-class="stub"1
3n-2
+class="stub"1
3n-2
-class="stub"1
3n+1
]
=class="stub"1
3
(1-class="stub"1
3n+1
)

class="stub"1
3

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