Question

已知数列an的前n项的和为sn,且对任意n∈N有an+sn=n,

Answer 1

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.