f(x)=(√2/2)(sinx+cosx)+3=sin(x+π/4)+3
sin(x+π/4)=1时,有f(x)max=4；sin(x+π/4)=-1时,有f(x)min=2
a1+b1=f(x)max a1-b1=f(x)min
a1+b1=4
a1-b1=2
解得a1=3 b1=1
q=S2/b2=S2/(b1q)=S2/(1×q)=S2/q
S2=q²
b2S2=b1qS2=1×q×q²=q³=27
q=3
S2=a1+a2=3+a2=q²=9
a2=6
d=a2-a1=6-3=3
an=a1+(n-1)d=3+3(n-1)=3n
bn=b1q^(n-1)=1×3^(n-1)=3^(n-1)
数列{an}的通项公式为an=3n；数列{bn}的通项公式为bn=3^(n-1)
Sn=(a1+an)n/2=(3+3n)n/2=(3/2)n(n+1)
1/Sn=(2/3)/[n(n+1)]=(2/3)[1/n -1/(n+1)]
1/bn=1/3^(n-1)
cn=1/Sn -1/bn
Tn=(2/3)[1/1-1/2+1/2-1/3+...+1/n -1/(n+1)] -1×(1-1/3ⁿ)/(1-1/3)
=(2/3)[1-1/(n+1)] -(3/2)(1-1/3ⁿ)
=1/[2×3^(n-1)] -2/[3(n+1)] -5/6