已知等差数列{an}的首项为a,公差为b,且不等式ax^2-3x+2>0的解集为(-无穷,1)∪(b,+无穷)若数列{b

由ax^2-3x+2>0解集为(-∞,1)U(b,+∞)得x=1时,ax^2-3x+2=0
a-3+2=0
a=1
x^2-3x+2>0
(x-1)(x-2)>0
x2
b=2
数列{an}是以1为首项,2为公差的等差数列.
an=1+2(n-1)=2n-1
bn=an×2ⁿ=(2n-1)×2ⁿ
Tn=b1+b2+...+bn=1×2+3×2^2+5×2^3+...+(2n-1)×2ⁿ
2Tn=1×2^2+3×2^3+...+(2n-3)×2ⁿ+(2n-1)×2^(n+1)
Tn-2Tn=-Tn=2+2×2^2+2×2^3+...+2×2ⁿ-(2n-1)×2^(n+1)
=2×(2+2^2+...+2ⁿ) -(2n-1)×2^(n+1) -2
=2×2×(2ⁿ-1)/(2-1)-(2n-1)×2^(n+1) -2
=(3-2n)×2^(n+1) -6
Tn=(2n-3)×2^(n+1) +6

由韦达定理知

1+b=3/a

1*b=2/a

得a=1 b=2

即an=2n-1

bn=(2n-1)*2^n

1. Tn=b1+b2+b3+b4+...+bn

2. 2Tn=2b1+2b2+2b3+...+2bn

由2-1得...