Question

老师，帮一下忙，第六题

Answer 1

（1）
(n，Sn)代入解析式得：
Sn=b^n+r
a1=S1=b+r
a1+a2=S2=b^2+r, a2=b^2-b
a1+a2+a3=b^3+r,a3=b^3-b^2=b(b^2-b)
因为{an}为等比数列，那么
a2/a1=a3/a2
所以(b^2-b)/(b+r)=b
==> -b=br
==> r=-1
(2)
b=2时，Sn=2^n-1
a1=1,公比q=a3/a2=b=2
an=2^(n-1)
bn=(n+1)/2^(n+1)
Tn=2/4+3/8+4/16+.....+(n+1)/2^(n+1) ①
两边同时乘以1/2
1/2Tn=2/8+3/16+4/32+....+n/2^（n+1)+(n+1)/2^(n+2) ②
①-②:
1/2Tn=1/2+1/8+1/16+....+1/2^(n+1)-(n+1)/2^(n+2)
=1/2+1/8*[1-1/2^(n-1)]/(1-1/2)-(n+1)/2^(n+2)
=1/2+1/4-1/2^(n+1)-(n+1)/2^(n+2)
=3/4-(n+3)/2^(n+2)
Tn=3/2-(n+3)/2^(n+1)