已知数列{an}满足,a1=1,a2=2,a(n+1)=a(n+2)+an,求a2011?
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已知数列{an}满足,a1=1,a2=2,a(n+1)=a(n+2)+an,求a2011?
已知数列{an}满足,a1=1,a2=2,a(n+1)=a(n+2)+an,求a2011?
已知数列{an}满足,a1=1,a2=2,a(n+1)=a(n+2)+an,求a2011?
a(n+2)=a(n+1)- an
a1=1,a2=2,a3=1,a4=-1,a5=-2,a6=-1 ,a7=1,a8,=2 ,...
这是一个周期为6的数列,
2011÷6=335.1
∴a2011 =a1 =1
由已知a(n 1)=an/a(n-1) 则a(n 2)=a(n 1)/an 两式相乘得,a(n 2)=1/a(n-1) 则a(n 5)=1/a(n 2)=a(n-1) 即a(n 6)=an
a(n+1)=a(n+2)+an-------1
所以
a(n+2)=a(n+1)+a(n+3)--------2
1 式加2式
a(n+1)+a(n+2)=a(n+2)+an+a(n+1)+a(n+3)
所以,0=an+a(n+3)
即an=-a(n+3)
所以a(n+3)=-a((n+3)+3)=-a(n+6)
所以an=-a(n+3)=-[-a(n+6)]=a(n+6)
即周期为6
所以a2011=a(6X335+1)=a1=1
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