Question

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

Answer 1

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