- 请赶快帮忙,一道数列问题,怎样用归纳法来解数列{an}满足a1=
- 数列{an}满足a1=0,an+1=(an-√3 )/(√3an+1),则a2009=
- 数列{an}满足a1=0,an+1=(an-√3 )/(√3an+1),则a2009=
解:a2=(a1-√3 )/(√3a1+1)=(0-√3 )/(0+1)=-√3
a3=(a2-√3 )/(√3a2+1)=(-√3-√3 )/(√3*(-√3)+1)=√3
a4=(a3-√3 )/(√3a3+1)=(√3-√3 )/(√3*√3)+1)=0
规律:
a1,a2,a3,a4,a5,a6,................................
0,-√3,√3,0,,-√3,√3,0,,-√3,√3,...............
a1=0,a4=0,a7=0,......,2008=0
∴a2009=-√3