已知数列{an}满足:1a1+1a2+1a3+…+1an=n2(n≥1,n∈N+),(1)求a2011(2)若bn=anan+1,Sn为数列{bn}的前b项和,存在正整数b,使得Sn>λ-12,求实数

题目简介

已知数列{an}满足:1a1+1a2+1a3+…+1an=n2(n≥1,n∈N+),(1)求a2011(2)若bn=anan+1,Sn为数列{bn}的前b项和,存在正整数b,使得Sn>λ-12,求实数

题目详情

已知数列{an}满足:
1
a1
+
1
a2
+
1
a3
+…+
1
an
=n2(n≥1,n∈N+),
(1)求a2011
(2)若bn=anan+1,Sn为数列{bn}的前b项和,存在正整数b,使得Sn>λ-
1
2
,求实数λ的取值范围.
题型:解答题难度:中档来源:不详

答案

(1)class="stub"1
a1
+class="stub"1
a2
+class="stub"1
a3
+…+class="stub"1
a2011
=20112
class="stub"1
a1
+class="stub"1
a2
+class="stub"1
a3
+…+class="stub"1
a2010
=20102
两式相减得class="stub"1
a2011
=20112-20102=4021⇒a2011=class="stub"1
4021

(2)class="stub"1
a1
+class="stub"1
a2
+class="stub"1
a3
+…+class="stub"1
an
=n2
class="stub"1
a1
+class="stub"1
a2
+class="stub"1
a3
+…+class="stub"1
an+1
=(n+1)2
两式相减得class="stub"1
an
=n2-(n-1)2=2n-1⇒an=class="stub"1
2n-1
(n≥2)
当n=1时,a1=1也满足上式∴an=class="stub"1
2n-1
(n≥1)
bn=anan+1=class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
class="stub"1
2n-1
-class="stub"1
2n+1

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

存在正整数b,使得Sn>λ-class="stub"1
2
,即Sn的最大值大于λ-class="stub"1
2

而Sn=class="stub"1
2
(1-class="stub"1
2n+1
)<class="stub"1
2

class="stub"1
2
>λ-class="stub"1
2
,即λ<1

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