设各项了均为正数的数列｛an｝的前n项和为Sn,已知a1=1,且（S（n+1）+λ）an=（Sn+1）a（n+1）对一切

(1)
(S(n+1)+λ)an=(Sn+1)a(n+1)
λ=1
(S(n+1)+1)an=(Sn+1)a(n+1)
n=1
(a1+a2+1)a1=(a1+1)a2
2+a2= 2a2
a2=2
(S(n+1)+1)an=(Sn+1)a(n+1)
a(n+1)/an = (S(n+1)+1)/(Sn+1)
=(S2+1)/(S1+1)
= 4/2
=2
an=2^(n-1) .a1
=2^(n-1)
(2)
a1=1
(S(n+1)+λ)an=(Sn+1)a(n+1)
n=1
(a1+a2+λ)a1=(a1+1)a2
1+a2+λ=2a2
a2=1+λ
n=2
(a1+a2+a3+λ)a2=(a1+a2+1)a3
(2+2λ+a3)(1+λ)=(3+λ)a3
(2+2λ)(1+λ) + (1+λ)a3 =(3+λ)a3
a3= 1+2λ+λ^2
{an}是等差数列
a3+a1 = 2a2
2+2λ+λ^2 = 2(1+λ)
λ^2=0
λ=0

1）若λ=1，则（S（n+1）+λ）an=（Sn+1）a（n+1）两边除以ana（n+1）得
S（n+1）/a（n+1）+1/a（n+1）=Sn/an+1/an
∴Sn/an+1/an，是常数列。Sn/an+1/an=2
解得，Sn=2an-1，∴an=2^（n-1）
（2）a...