已知数列{an}满足a1=1,a2=-13,an+2-2an+1+an=2n-6b1=a2-a1=-13-1=-14bn=a(n+1)-ana(n+2)-2a(n+1)+an=2n-6a(n+2)-a(n+1)-a(n+1)+an=2n-6a(n+2)-a(n+1)-[a(n+1)+an]=2n-6b(n+1)-bn=2n-6所以bn-b(n-1)=2(n-1)-6bn-b(n-1)=2(n-1)-6.b3-b2=2*2-6b2-b1=2*1-6以上等

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