茜岚
幼苗
共回答了19个问题采纳率:94.7% 举报
∵知函数f(x)=
1
x-1 ,各项均为正数的数列{a n }满足a n+2 =f(a n ),
∴a n+2 =
1
a n -1 ,
取n=2011,a 2011 =a 2013 ,a n+2 =
1
a n -1 ,
可得a 2013 =
1
a 2011 -1 =a 2011 ,所以(a 2011 ) 2 -a 2011 -1=0,
∴a 2011 是方程x 2 -x-1=0的根,a 2011 >0
∴a 2011 =
5 +1
2 ,
∵a n+2 =
1
a n -1 ,
∴a 2009 =
1
a 2011-1 =
1
5 +1
2 -1 =
2(
5 +1)
4 =
5 +1
2 ,
a 2007 =
1
a 2009 -1 =
5 +1
2
a 2006 =
1
a 2007 -1 =
5 +1
2
依此类推可得
∴a 1 =
1
a 2 -1 =
5 +1
2
故答案为:
5 +1
2 ;
1年前
5