∵Sn=1/3(An-1)
∴Sn-1=1/3(An-1 -1)
∴An=Sn-Sn-1=1/3(An-1)-1/3(An-1 -1)
化简得2/3An=-1/3An-1
即An/An-1=-1/2=q
∴数列为等比数列
a1=s1=1/3(a1-1)
a1=-1/2.
a2=s2-a1=1/3(a2-1)+1/2=1/3a2+1/6
a2=1/4
所以an=(-1/2)*(-1/2)^(n-1)
(1)(a+d)^2=(a)*(a+3d)
a=d
a+2d+a+4d=8
a=d=1
(2)bn=2^(-n+1)
an=n
cn=n*2^(-n+1)=n/(2^(n-1))
Tn=1/1+2/2+3/4+…+n/(2^(n-1))
(1/2)*Tn=1/2+2/4+3/8+……+(n-1)/(2^(n-1))+n/(2^n)
(1/2)*Tn=1+1/2+1/4+……+1/(2^(n-1))-n/(2^n)=2-1/（2^(n-1)）-n/(2^n)=2-(2+n)/(2^n)
Tn=4-(4+2n)/(2^n)
(3)显然Tn<4
对（4+2n）/(2^n)求导发现其为单调递减
n=1时Tn=1>0
m=0,M=4