已知数列{an}中,a1=2,an=a(n-1)+1,数列{bn}的通项公式为bn=2^(n+1),数列{an·bn}的前n项的和为Tn

an=a(n-1)+1
an -a(n-1) =1
an -a1=n-1
an= n+1

let
S =1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S= n.2^(n+1)- [ 2+2^2+...+2^n]
=n.2^(n+1)- 2(2^n -1)
cn= an.bn
=(n+1).2^(n+1)
= 2(n.2^n) + 2^(n+1)
Tn = c1+c2+...+cn
=2S + 4(2^n-1)
=2n.2^(n+1)
=4n.2^n
(2)
2n·4^n-(-1)^(n+1)λTn>0

2n·4^n-(-1)^(n+1)λ.(4n.2^n)>0
2n·2^n . [ 2^n -(-1)^(n+1).2λ ] >0
2^n -(-1)^(n+1).2λ >0
2^n >2λ
λ <1