已知数列{an}满足a1=1,a2=-13,a(n+2)-2a(n+1)+an=2n-6

a(n+2)-2a(n+1)+an=2n-6,
[a(n+2)-a(n+1)]-[a(n+1)-an]=2n-6,
令bn=a（n+1）-an,则b(n+1)-bn=2n-6,
b2-b1=-4
b3-b2=-2,
...
bn-b(n-1)=2(n-1)-6,
相加得bn-b1=2n(n-1)/2 -6(n-1)=n^2-7n+6
b1=a2-a1=-14,所以bn=n^2-7n-8
an最小就是an-a(n-1)=b(n-1)不再为负数,
bn=n^2-7n-8>=0算出n>=8,n=8时刚好b8=0,
即a9-a8=0,所以n=8或9时最小