已知数列{an}是公差为2的等差数列,且a1+1,a3+1,a7+1成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)令bn=1a2n-1(n∈N*),记数列{bn}的前n项和为Tn,求证:Tn<14

题目简介

已知数列{an}是公差为2的等差数列,且a1+1,a3+1,a7+1成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)令bn=1a2n-1(n∈N*),记数列{bn}的前n项和为Tn,求证:Tn<14

题目详情

已知数列{an}是公差为2的等差数列,且a1+1,a3+1,a7+1成等比数列.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)令bn=
1
a2n
-1
(n∈N*)
,记数列{bn}的前n项和为Tn,求证:Tn
1
4
题型:解答题难度:中档来源:成都模拟

答案

(I)设等差数列{an}的首项为a1,因为a1+1,a3+1,a7+1成等比数列,所以有
         (a3+1)2=(a1+1)(a7+1),即(a1+5)2=(a1+1)(a1+13),
        解得:a1=3,所以an=3+2(n-1)=2n+1;
(II)证明:由(I)知:an=2n+1,所以
      bn=class="stub"1
an2-1
=class="stub"1
(2n+1)2-1
=class="stub"1
4
•class="stub"1
n(n+1)
=class="stub"1
4
(class="stub"1
n
-class="stub"1
n+1
)

所以Tn=class="stub"1
4
(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
)
=class="stub"1
4
(1-class="stub"1
n+1
)
=class="stub"1
4
-class="stub"1
4(n+1)
<class="stub"1
4

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