设等比数列{an}公比为q.
x=1 y=S1=a1,x=2,y=S2=a1+a2,x=3,y=S3=a1+a2+a3分别代入
a1=6+r (1)
a1+a2=12+r (2)
a1+a2+a3=24+r (3)
(2)-(1)
a2=6
(3)-(1)
a2+a3=18
a3=18-a2=18-6=12
q=a3/a2=12/6=2
a1=a2/q=6/2=3
r=a1-6=3-6=-3
数列{an}的通项公式为an=3×2^(n-1)=(3/2)2^n
bn=3n/an=2n/2^n
Tn=2(1/2^1+2/2^2+3/2^3+...+n/2^n)
Tn/2=2[1/2^2+2/2^3+...+(n-1)/2^n+n/2^(n+1)]
Tn-Tn/2=Tn/2=2(1/2^1+1/2^2+...+1/2^n-n/2^(n+1)]
Tn=4[1/2^1+1/2^2+...+1/2^n-n/2^(n+1)]
=4[(1/2)(1-1/2^n)/(1-1/2)]-2n/2^n
=4-4/2^n-2n/2^n
=4-(2n+4)/2^n
=4-(n+2)/2^(n-1)