题目内容
函数f(x)由表定义:若a0=5,an+1=f(an),n=0,1,2,…,则a2009=______
| x | 2 | 5 | 3 | 1 | 4 |
| f(x) | 1 | 2 | 3 | 4 | 5 |
∵a0=5,an+1=f(an),n=0,1,2,
∴a1=f(a0)=f(5)=2;
a2=f(a1)=f(2)=1;
a3=f(a2)=f(1)=4;
a4=f(a3)=f(4)=5;
a5=f(a4)=f(5)=2;
∴递推数列a0=5,an+1=f(an)是周期为4的数列.
∴a2009=f(a2008)=f(5)=2.
故答案为:2
∴a1=f(a0)=f(5)=2;
a2=f(a1)=f(2)=1;
a3=f(a2)=f(1)=4;
a4=f(a3)=f(4)=5;
a5=f(a4)=f(5)=2;
∴递推数列a0=5,an+1=f(an)是周期为4的数列.
∴a2009=f(a2008)=f(5)=2.
故答案为:2
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函数f(x)由表定义:若a=5,an+1=f(an),n=0,1,2,…,则a2009=
| x | 2 | 5 | 3 | 1 | 4 |
| f(x) | 1 | 2 | 3 | 4 | 5 |
函数f(x)由表定义:若a=5,an+1=f(an),n=0,1,2,…,则a2009=
| x | 2 | 5 | 3 | 1 | 4 |
| f(x) | 1 | 2 | 3 | 4 | 5 |