a1=b b为常数， an=2a(n-1)-2^(n-1)

(1)
an=2a(n-1)-2^(n-1)
an/2^n=(2a(n-1))/2^n-(2^(n-1)/2^n)
=(a(n-1)/2^(n-1))-(1/2)
(an/2^n)-(a(n-1)/2^(n-1))=-1/2
所以:{an/2^n}为等差数列,公差d=-1/2

(2)设an/2^n=cn
c1=a1/2=b/2
cn=c1+(n-1)d=(b/2)+(n-1)(-1/2)=(b+1-n)/2
an=cn*2^n=(b+1-n)*2^(n-1)

(3)
an=(b+1-n)*2^(n-1)=(b+1)*2^(n-1)-n*2^(n-1)
设pn=(b+1)*2^(n-1), wn= n*2^(n-1), pn,wn的前n项和Pn, Wn
则：an=pn-wn
w(n+1)=(n+1)*2^n
(1/2)w(n+1)=(n+1)*2^(n-1)
(1/2)w(n+1)-wn=2^(n-1)
(1/2)(w(n+1)+Wn-w1)-Wn=1+2+2^2+…+2^(n-1)=(1-2^n)/(1-2)=2^n-1
Wn=w(n+1)-w1-2*2^n+2=(n+1)*2^n-1-2*2^n+2
=(n-1)*2^n+1
Pn=(b+1)(1+2+2^2+…+2^(n-1)
=(b+1)(1-2^n)/(1-2)=(b+1)(2^n-1)
所以:Sn=Pn-Wn=(b+1)(2^n-1)- ((n-1)*2^n+1)
=(b+2-n)*2^n-b-2