已知数列{an}满足:a1=2,且an+1=2-1an,n∈N*.(1)设bn=1an-1,求证:{bn}是等差数列;(2)求数列{an}的通项公式;(3)设cn=an+1an,求证:2n<c1+c2

题目简介

已知数列{an}满足:a1=2,且an+1=2-1an,n∈N*.(1)设bn=1an-1,求证:{bn}是等差数列;(2)求数列{an}的通项公式;(3)设cn=an+1an,求证:2n<c1+c2

题目详情

已知数列{an}满足:a1=2,且an+1=2-
1
an
,n∈N*
(1)设bn=
1
an-1
,求证:{bn}是等差数列;
(2)求数列{an}的通项公式;
(3)设cn=an+
1
an
,求证:2n<c1+c2+…+cn<2n+1,n∈N*
题型:解答题难度:中档来源:不详

答案

(1)∵a1=2,且an+1=2-class="stub"1
an
,n∈N*.
∴a2=2-class="stub"1
2
=class="stub"3
2

a3=2-class="stub"2
3
=class="stub"4
3

a4=1-class="stub"3
4
=class="stub"5
4


猜想an=class="stub"n+1
n

用数学归纳法进行证明:
a1=class="stub"2
1
=2
,成立.
②假设n=k时,成立,即ak=class="stub"k+1
k

则当n=k+1时,ak+1=2-class="stub"1
ak
=2-class="stub"k
k+1
=class="stub"k+2
k+1
,成立.
由①②知,an=class="stub"n+1
n

∵bn=class="stub"1
an-1

∴bn+1-bn=class="stub"1
an+1-1
-class="stub"1
an-1

=class="stub"1
1-class="stub"1
an
-class="stub"1
1-class="stub"1
an-1

=class="stub"1
1-class="stub"n
n+1
-class="stub"1
1-class="stub"n-1
n

=(n+1)-n=1,
∴数列{bn}是等差数列.
(2))∵a1=2,且an+1=2-class="stub"1
an
,n∈N*.
∴a2=2-class="stub"1
2
=class="stub"3
2

a3=2-class="stub"2
3
=class="stub"4
3

a4=1-class="stub"3
4
=class="stub"5
4


猜想an=class="stub"n+1
n

用数学归纳法进行证明:
a1=class="stub"2
1
=2
,成立.
②假设n=k时,成立,即ak=class="stub"k+1
k

则当n=k+1时,ak+1=2-class="stub"1
ak
=2-class="stub"k
k+1
=class="stub"k+2
k+1
,成立.
由①②知,an=class="stub"n+1
n

(3)∵cn=an+class="stub"1
an
an=class="stub"n+1
n

cn=class="stub"n+1
n
+class="stub"n
n+1
= 2+class="stub"1
n
 -class="stub"1
n+1

∴c1+c2+…+cn=2n+(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1

=2n+1-class="stub"1
n+1
<2n+1.
∵c1+c2+…+cn=2n+(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1

=2n+1-class="stub"1
n+1
=2n+class="stub"n
n+1
>2n.
∴2n<c1+c2+…+cn<2n+1,n∈N*.

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