已知各项均不相等的等差数列{an}的前四项和S4=14,且a1,a3,a7成等比数列.(I)求数列{an}的通项公式;(II)设Tn为数列{1anan+1}的前n项和,若Tn≤λan+1对∀n∈N*恒

题目简介

已知各项均不相等的等差数列{an}的前四项和S4=14,且a1,a3,a7成等比数列.(I)求数列{an}的通项公式;(II)设Tn为数列{1anan+1}的前n项和,若Tn≤λan+1对∀n∈N*恒

题目详情

已知各项均不相等的等差数列{an}的前四项和S4=14,且a1,a3,a7成等比数列.
(I)求数列{an}的通项公式;
(II)设Tn为数列{
1
anan+1
}的前n项和,若Tn≤λan+1对∀n∈N*恒成立,求实数λ的最小值.
题型:解答题难度:中档来源:不详

答案

(I)设公差为d,由已知得:
S4=14
a32=a1a7

4a1+class="stub"4×3
2
d=14
(a1+2d)2=a1(a1+6d)

解得:d=1或d=0(舍去),
∴a1=2,
故an=2+(n-1)=n+1;
(II)∵class="stub"1
anan+1
=class="stub"1
(n+1)(n+2)
=class="stub"1
n+1
-class="stub"1
n+2

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

∵Tn≤λan+1对∀n∈N*恒成立,即class="stub"n
2(n+2)
≤λ(n+2),λ≥class="stub"n
2(n+2)2
∀n∈N*恒成立,
class="stub"n
2(n+2)2
=class="stub"1
2(n+class="stub"4
n
+4)
class="stub"1
2(4+4)
=class="stub"1
16

∴λ的最小值为class="stub"1
16

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