已知数列{an}的通项公式an与前n项Sn公式之间满足Sn=2-3an,1求a1求an与an的递推关系n≥2,n∈N*求sn与sn的递推关系(n≥2,n∈N*

Question

已知数列{an}的通项公式an与前n项Sn公式之间满足Sn=2-3an,
1求a1
求an与an的递推关系n≥2,n∈N*
求sn与sn的递推关系(n≥2,n∈N*

Answer 1

a(1)=s(1)=2-3a(1),a(1) = 1/2.
s(n) = 2 - 3a(n),
s(n+1) = 2-3a(n+1),
a(n+1)=s(n+1)-s(n) = 3a(n) - 3a(n+1),
a(n+1) = (3/4)a(n),
{a(n)} 是首项为a(1)=1/2,公比为(3/4)的等比数列.
a(n) = (1/2)(3/4)^(n-1)
s(n+1) = 2 - 3a(n+1) = 2 - 3[s(n+1)-s(n)],
4s(n+1) = 3s(n) + 2,
s(n+1) = (3/4)s(n) + 1/2