已知数列an的前n项的和为sn,且对任意n∈N有an+sn=n,

1=a(1)+s(1)=2a(1),a(1)=1/2=s(1).
a(n)+s(n)=n,
a(n+1)+s(n+1)=n+1,
1 = (n+1)-n = a(n+1)-a(n) + [s(n+1)-s(n)] = 2a(n+1)-a(n),
2a(n+1) = a(n) + 1,
2a(n+1)-2 = 2[a(n+1)-1] = a(n)-1,
a(n+1)-1 = [a(n)-1]/2,
{b(n)=a(n)-1}是首项为b(1)=a(1)-1=-1/2,公比为（1/2）的等比数列.
a(n)-1 = (-1/2)(1/2)^(n-1) = -1/2^n,
a(n) = 1 - 1/2^n,
c(1)=a(1)=1/2.
c(n+1)=a(n+1)-a(n)=[1-1/2^(n+1)] - [1-1/2^n] = 1/2^(n+1),
c(n) = 1/2^n.

1、
an+Sn=n
a（n+1）+Sn+a（n+1）=n+1
两式相减得2a（n+1）-an=1
整理变形得2（a（n+1）-1）=an-1
可知，bn=an-1为等比数列，等比为1/2

2、取n=1，得2a1=1，则a1=1/2
结合第一问答案，可得an-1=（-1/2)*(1/2)^(n-1)=(-1/2^)n
则a...