已知数列{an}满足a1=2,an=(1⼀2)an+1-2^n(n갠N*)

1、

an=(1/2)an+1 - 2^n 即 an + 2^n =(1/2)an+1
得
2an+2^n+1=an+1 两边同时除于2^n+1得

(an/2^n) +1=(an+1)/(2^n+1) 所以数列{an/2^n}是首项为1，公差为1等差数列

2、

设an/2^n=bn
则bn=n 所以an=n x 2^n

sn= 1x2+2x2^2+3x2^3+....+nx2^n

2sn=1x2^2+2x2^3+3x2^4+.....+nX2^n+1

2sn-sn= -- (2+2^2+2^3+….+2^n)+nx2^n+1=sn

所以sn= nx2^(n+1)
--- 2^(n+1) + 2 =(n-1)x2^(n+1) + 2

(1)
an=(1/2)a(n+1)-2^n
an/2^n = a(n+1)/2^(n+1) - 1/2
a(n+1)/2^(n+1) - an/2^n =1/2
=>{an/2^n}是等差数列

(2)
a(n+1)/2^(n+1) - an/2^n =1/2
an/2^n - a1/2^1 = (n-1)/2
an/2^n = n/2
an = n.2^(n-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) = n.2^(n+1) -(n+1).2^n +1
= (n-1). 2^n +1
Sn = a1+a2+..+an
=∑(i:1->n) i.2^(i-1)
= (n-1). 2^n +1

第一个同除2的n次幂，第二个把an的通项求出来，再算，Sn表达式应该是一个等比数列加一个等差数列。首项为1 公差为1 -Sn=[2(1-2^n)]/(1-2)-n*2^(n+1)

解:(1)因为an=(1/2)a(n＋1)-2^n ,所以a(n＋1)=2an＋2^(n＋1),所以a(n＋1)/2^(n＋1)-an/2n=[2an＋2^(n＋1)]/2^(n＋1)-an/2^n=an/2^n＋1-an/2^n=1,所以数列{an/2^n}是首项为2,公差为1的等差数列。