Sn=f(n)=an²－9n,则：
a1=S1=a－9,
当n≥2时,an=Sn－S(n－1)=f(n)－f(n－1)=2an－a－9,则公差d=an－a(n－1)=2a=2,得：a=1
所以f(x)=x²－9x,an=2n－10,Sn=n²－9n
17、f(x)=4sin(wx)sin²[(wx/2)＋π/4]＋cos2wx
=2sin(wx)[1－cos(wx＋π/2)]＋cos2wx
=2sin(wx)[1＋sin(wx]＋cos(2wx)
=2sin(wx)＋2sin²(wx)＋cos(2wx)
=2sin(wx)＋1－cos2wx＋cos(2wx)
=2sin(wx)＋1
①若w=1,则：f(x)=2sinx＋1,则f(x)的最小正周期是2π/1=2π
②若f(x)在[－π/2,2π/3]上递增,则函数f(x)的半个周期T/2应该满足：
T/2≥2×(2π/3)
T≥8π/3
2π/w≥8π/3
得：0