数列an满足,a1=1/4,a2=3/4,an+1=2an-an-1(n≥2,n属于N*),数列bn满足b1

(1)∵a(n+1)=2an-a(n-1)
∴2an=a(n+1)+a(n-1).等差中项的性质
∴﹛an﹜是等差数列
(2)a1=1/4,a2=3/4
an=1/4+(n-1)×(3/4-1/4)=n/2-1/4
∵3bn-b(n-1)=n
∴bn=b(n-1)/3+n/3 (n≥2)
∴b(n+1)-a(n+1)=1/3bn+(n+1)/3-(n+1)/2+1/4
= 1/3bn-n/6+1/12
=1/3(bn-n/2+1/4)
=1/3(bn-an)
∴﹛bn-an﹜等比数列
(3)bn-an=(b1-1/4)(1/3)^(n-1)
∴bn=(b1-1/4)(1/3)^(n-1)+n/2-1/4
当n≥2时,bn-b(n-1)=(b1-1/4)[(1/3)^(n-1)-(1/3)^(n-2)]+1/2
=1/2-2/3(b1-1/4)(1/3)^(n-2)
∵b1＜0
∴bn-b(n-1)＞0
∴﹛bn﹜是单增数列
Sn=b1+b2+..+bn
∵当且仅当n=4,Sn取最小
∴b4＜0,b5＞0
∴(b1-1/4)(1/3)^3+4/2-1/4
=（b1-1/4)/27+7/4＜0
∴b1＜-47
(b1-1/4)(1/3)^4+5/2-1/4
=（b1-1/4)/81+9/4＞0
∴b1＞-182
∴-182＜b1＜-47