题目内容
数列{an}满足a1>1,an+1-1=an(an-1),(n∈N+),且
+
+…+
=2,则a2013-4a1的最小值为______.
1 |
a1 |
1 |
a2 |
1 |
a2012 |
a1>1,由an+1-1=an(an-1),(n∈N+)知,对所有n,an>1,
等式两边取倒数,得
=
=
-
,得,
=
-
,
则
+
+…+
=
-
=2
整理可得,a2013=
,
a2013-4a1=2(3-2a1)+
-
≥2
-
=-
.
则a2013-4a1的最小值为 -
.
故答案为:-
.
等式两边取倒数,得
1 |
an+1-1 |
1 |
an(an-1) |
1 |
an-1 |
1 |
an |
1 |
an |
1 |
an-1 |
1 |
an+1-1 |
则
1 |
a1 |
1 |
a2 |
1 |
a2012 |
1 |
a1 |
1 |
a2013 |
整理可得,a2013=
2-a1 |
3-2a1 |
a2013-4a1=2(3-2a1)+
1 |
2(3-2a1) |
11 |
2 |
(3-2a1)
|
11 |
2 |
7 |
2 |
则a2013-4a1的最小值为 -
7 |
2 |
故答案为:-
7 |
2 |
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