已知数列﹛an﹜是一个公差大于零的等差数列,且a3a6=55,a2+a7=16,数列﹛bn﹜的前n项和为sn,且sn=2bn-2

1.
设公差为d,则d>0
数列是等差数列,a3+a6=a2+a7=16,又a3a6=55,a3、a6是方程x²-16x+55=0的两根.
(x-5)(x-11)=0
x=5或x=11
d>0 a6>a3
a3=5 a6=11
a6-a3=3d=11-5=6
d=2
an=a1+(n-1)d=a3+(n-3)d=5+2(n-3)=2n-1
n=1时,b1=S1=2b1-2
b1=2
n≥2时,bn=Sn-S(n-1)=2bn-2-[2b(n-1)-2]
bn=2b(n-1)
bn/b(n-1)=2,为定值,数列{bn}是以2为首项,2为公比的等比数列.bn=2ⁿ
数列{an}的通项公式为an=2n-1；数列{bn}的通项公式为bn=2ⁿ
2.
cn=an/bn=(2n-1)/2ⁿ
Tn=c1+c2+c3+...+cn=1/2+3/2²+5/2³+...+(2n-1)/2ⁿ
Tn /2=1/2²+3/2³+...+(2n-3)/2ⁿ+(2n-1)/2^(n+1)
Tn-Tn /2=Tn /2=1/2 +2/2²+2/2³+...+2/2ⁿ-(2n-1)/2^(n+1)
Tn=1+2/2+2/2²+...+2/2^(n-1) -(2n-1)/2ⁿ
=2[1+1/2+1/2²+...+1/2^(n-1)] -(2n-1)/2ⁿ -1
=2×1×(1-1/2ⁿ)/(1-1/2) -(2n-1)/2ⁿ -1
=3 -(2n+3)/2ⁿ