已知数列{an}满足a1=a,a2=b,a(n+1)=a(n+2)+an,求a2012

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已知数列{an}满足a1=a,a2=b,a(n+1)=a(n+2)+an,求a2012

已知数列{an}满足a1=a,a2=b,a(n+1)=a(n+2)+an,求a2012
已知数列{an}满足a1=a,a2=b,a(n+1)=a(n+2)+an,求a2012

已知数列{an}满足a1=a,a2=b,a(n+1)=a(n+2)+an,求a2012
由a(n+1)=a(n+2)+an
易知:
a1=a
a2=b
a3=a2-a1=b-a
a4=a3-a2=-a
a5=a4-a3=-b
a6=a5-a4=a-b
a7=a6-a5=a.=a1
a8=a7-a6=b.=a2
a9=a8-a7=b-a.=a3
……
故an=a,n=6k+1,k∈N+
……..b,n=6k+2,k∈N+
……..b-a,n=6k+3,k∈N+
……..-a,n=6k+4,k∈N+
……..-b,n=6k+5,k∈N+
……..a-b,n=6k,k∈N+
∵2012=6×335+2
∴a(2012)=b

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