Question

设等比数列An的前n项和为Sn，对任意正整数n，都有An+1=2Sn-1，求通项公式An

Answer 1

a(n+1) =2S(n-1) (1)
a(n)=2S(n-2) (2)
a(n+1)-an=2a(n-1)
a(n+2)-a(n+1)-2an=0
The auxilary equation
x^2-x-2=0
(x-2)(x+1)=0
x=2 or -1
an= A(2)^(n-1)+B(-1)^(n-1)
a1 = A+B (3)
a2=2A-B (4)
(3)-(4)
3A=a1+a2
A= (a1+a2)/3
B=(2a1-a2)/3
ie
an=[(a1+a2)/3](2)^(n-1) +[(2a1-a2)/3](-1)^(n-1)