数列{an}中,已知a1=5/6,an+1=1/3an+(1/2)^n+1,(n∈N*),
问题描述：
数列{an}中,已知a1=5/6,an+1=1/3an+(1/2)^n+1,(n∈N*),
且数列a2-1/2a1,a3-1/2a2.an-1/2an-1..是公比为1/3的等比数列,求数列{an}的通项公式.
答
a(n+1)=(1/3)an+(1/2)^(n+1)
a(n+1)-3(1/2)^(n+1) = (1/3)[an - 3(1/2)^n ]
{an - 3(1/2)^n} 是等比数列,q= 1/3
an - 3(1/2)^n = (1/3)^(n-1) .{a1 - 3(1/2)^1}
= -2(1/3)^n
an =3(1/2)^n - 2(1/3)^n