Question

已知数列{an}的前n项和为Sn，对任意的n∈N*,点(an,Sn)都在直线2x-y-2=0的图像上

(1)求{an}的通项公式
(2)是否存在等差数列{bn},使得a1b1+a2b2+…+anbn=(n-1)·2^n+1+2对一切n∈N*都成立？若存在，求出{bn}的通项公式

Answer 1

(1)
点(an,Sn)都在直线2x-y-2=0的图像上
2an-Sn-2=0
n=1, => a1=2

2an-Sn-2 =0
2(Sn-S(n-1)-Sn -2 =0
Sn+2 = 2[S(n-1)+2]
(Sn+2)/[S(n-1)+2]=2
(Sn+2)/(S1+2)=2^(n-1)
Sn = -2+ 2^(n+1)
an = Sn-S(n-1)
= 2^n
(2)
bn = b1+(n-1)d
anbn =(b1+(n-1)d).2^n
= (b1-d).2^n + 2d[n.2^(n-1)]
a1b1+a2b2+...+anbn
= 2(b1-d)(2^n-1) + 2d[∑(i:1->n) i.2^(i-1) ]

consider
1+x+x^2+...+x^n = (x^(n+1)-1)/(x-1)
1+2x+....+nx^(n-1) = [(x^(n+1)-1)/(x-1)]'
= [nx^(n+1)-(n+1)x^n+1]/(x-1)^2
put x=2
[∑(i:1->n) i.2^(i-1) ] = 1+ (n-1).2^n

a1b1+a2b2+...+anbn
= 2(b1-d)(2^n-1) + 2d[ 1+ (n-1).2^n ]
=[b1-d +(n-1)d] .2^(n+1) + (4d-2b1)
= (b1-d).2^(n+1) + (n-1)d.2^(n+1) + (4d-2b1)

=>(b1-d).2^(n+1) + (n-1)d.2^(n+1) + (4d-2b1) = (n-1)·2^(n+1)+2
coef. of n.2^(n+1)
d=1

coef. of constant
4d-2b1=2
b1=1

bn = b1+(n-1)d
= 1+ (n-1)1
= n