设数列{a(n)}的前n项和为Sn,已知ba(n)-2^n=(b-1)Sn求{a(n)}的通项公式

ba(n)-2^n = (b-1)S(n), ba(1) - 2 = (b-1)S(1) = (b-1)a(1), a(1)=2.
ba(n+1)-2^(n+1)=(b-1)S(n+1),
ba(n+1)-2^(n+1)-ba(n)+2^n = (b-1)[S(n+1)-S(n)] = (b-1)a(n+1),
a(n+1) = ba(n) + 2^n,
a(n+1)/2^n = 2ba(n)/2^(n-1) + 1,
c(n) = a(n)/2^(n-1),
c(n+1) = 2bc(n) + 1,
c(n+1) + x = 2b[c(n) + x], 1 = x(2b-1).
b = 1/2时，c(n+1)=c(n)+1,{c(n)}是首项为c(1)=a(1)/1=a(1)=2,公差为1的等差数列。c(n)=2+(n-1)=n+1.a(n)=c(n)2^(n-1)=(n+1)2^(n-1),n=1,2,...

b不等于1/2时，x = 1/(2b-1),
c(n+1) + 1/(2b-1) = 2b[c(n) + 1/(2b-1)],
{c(n) + 1/(2b-1)}是首项为c(1)+1/(2b-1)=2+1/(2b-1)=(4b-1)/(2b-1),公比为2b的等比数列。
c(n)+1/(2b-1) = (4b-1)/(2b-1)*(2b)^(n-1),
a(n)/2^(n-1) = c(n) = (4b-1)(2b)^(n-1)/(2b-1) - 1/(2b-1) = [(4b-1)(2b)^(n-1) - 1]/(2b-1),
a(n) = 2^(n-1)[(4b-1)(2b)^(n-1) - 1]/(2b-1), n = 1,2,...