Question

已知数列｛an｝的前n项和Sn满足：Sn=2(an-1),数列｛bn｝满足：对任意n∈N+有a1b1+a2b2+...+anbn=(n-1).2^

已知数列｛an｝的前n项和Sn满足：Sn=2(an-1),数列｛bn｝满足：对任意n∈N+有a1b1+a2b2+...+anbn=(n-1).2^（n+1）+2 (1)求数列｛an｝与数列｛bn｝的通项公式（2）记cn=bn/an，数列｛cn｝的前n项和为Tn，证明：当n≥6时，n|2-Tn|＜1.请详细回答，谢谢

Answer 1

Sn = 2(an-1)
n=1, a1=2
for n≥2
an = Sn - S(n-1)
=2an -2a(n-1)
an = 2a(n-1)
=2^(n-1) .a1
=2^n

a1.b1+a2.b2+...+an.bn=(n-1).2^(n+1)+ 2 (1)
n=1
a1.b1=2

b1=1

a1.b1+a2.b2+...+a(n-1).b(n-1)=(n-2).2^n + 2 (2)

(1)-(2)

an.bn = n2^n
2^n.bn =n.2^n
bn =n

(2)
记cn=bn/an，数列｛cn｝的前n项和为Tn
cn = n/2^n
=n.2^(-n)
let
S= 1.2^(-1)+2.2^(-2)+....+n.2^(-n) (3)
(1/2)S= 1.2^(-2)+2.2^(-3)+....+n.2^(-n-1) (4)
(3)-(4)
(1/2)S = ( 1/2 +1/2^2+...+1/2^n) - n.2^(-n-1)
= [ 1 - (1/2)^n ] - n.(1/2)^(n+1)
S = 2[ 1 - (1/2)^n ] - n(1/2)^n
=2 - (n+2).(1/2)^n
Tn = c1+c2+...+cn
=S
=2 - (n+2).(1/2)^n

n|2-Tn|
=n| (n+2).(1/2)^n |
=n(n+2).(1/2)^n
<1 (n≥6)