数列{An}中,A1=8,A4=,且满足：2A(n+2)-2A(n+1)+An=0

(1)a(n+2)-2a(n+1)+an=0
a(n+1)-a(n+1)=a(n+1)-an
所以数列an为等差数列
设an的公差为d
a4=a1+3d
d=-2
an=a1+(n-1)d
=8+2-2n
=10-2n
(2)bn=1/n(12-an)=1/n(2+2n)=2/(n+1)n
1/(n+1)n=1/n-1/(n+1)
Sn=b1+b2+b3+.+bn
=2/2*1+2/3*2+...+2/(n+1)n
=2[1-1/2+1/2-1/3+1/3-1/4+.+1/(n-1)-1/n+1/n-1/(n+1)]
=2[1-1/(n+1)]
因为1/(n+1)在定义域内为单调递减函数,当n=1时取最大值
所以当n=1时,Sn取最小值
2[1-1/2]