已知数列{an}、{bn}满足:a1=14,bn+1=bn1-an2an+bn=1.(1)求证:数列{1bn-1}是等差数列;(2)求数列{an}的通项公式;(3)设sn=a1a2+a2a3+a3a4

题目简介

已知数列{an}、{bn}满足:a1=14,bn+1=bn1-an2an+bn=1.(1)求证:数列{1bn-1}是等差数列;(2)求数列{an}的通项公式;(3)设sn=a1a2+a2a3+a3a4

题目详情

已知数列{an}、{bn}满足:a1=
1
4
bn+1=
bn
1-
an2
an+bn=1.
(1)求证:数列{
1
bn-1
}是等差数列;
(2)求数列{an}的通项公式;
(3)设sn=a1a2+a2a3+a3a4+…anan+1,若4aSn<bn对于n∈N*恒成立,试求实数a的取值范围.
题型:解答题难度:中档来源:不详

答案

(1)由an+bn=1,得bn=1-an,
依题意bn+1=
bn
1-
a2n
=
1-an
(1-an)(1+an)
=class="stub"1
1+an
class="stub"1
bn+1-1
-class="stub"1
bn-1
=class="stub"1
class="stub"1
1+an
-1
-class="stub"1
1-an-1
=-class="stub"1
an
-1+class="stub"1
an
=-1
a1=class="stub"1
4
,∴b1=class="stub"3
4
,class="stub"1
b1-1
=-4
,∴数列{class="stub"1
bn-1
}
是以-4为首项公差为-1的等差数列
(2)由(1)知class="stub"1
bn-1
=-4+(n-1)(-1)=-n-3

bn=-class="stub"1
n+3
+1=class="stub"n+2
n+3
an=1-bn=1-class="stub"n+2
n+3
=class="stub"1
n+3

(3)Sn=a1a2+a2a3+…+anan+1=class="stub"1
4×5
+class="stub"1
5×6
+class="stub"1
(n+3)•(n+4)
=class="stub"1
4
-class="stub"1
5
+class="stub"1
5
-class="stub"1
6
+class="stub"1
n+3
-class="stub"1
n+4
=class="stub"1
4
-class="stub"1
n+4
=class="stub"n
4(n+4)
∴4aSn-bn=class="stub"an
n+4
-class="stub"n+2
n+3
=
(a-1)n2+(3a-6)n-8
(n+3)(n+4)

依题意可知(a-1)n2+(3a-6)n-8<0恒成立,令f(n)=(a-1)n2+(3a-6)n-8
当a=1时,f(n)=-3n-8<0恒成立
当a>1时,由二次函数性质知f(n)<0不可能成立
当a<1时,此二次函数的对称轴为x=-class="stub"3a-6
2(a-1)
=-class="stub"3
2
(1-class="stub"1
a-1
)<0

则f(n)在n∈N*上是单调递减,∴要使f(n)<0对n∈N*恒成立
必须且只须f(1)<0即4a-15<0,∴a<class="stub"15
4
,又a<1∴a<1
综上a≤1,4aSn≤bn对于n∈N*恒成立.

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