若数列an的前n项和为Sn=2/3an+1/3,则数列an的通项公式是an=?
问题描述：
若数列an的前n项和为Sn=2/3an+1/3,则数列an的通项公式是an=?
答
a(1)=s(1)=(2/3)a(1)+1/3,a(1)=1.s(n) = (2/3)a(n) + 1/3,s(n+1)=(2/3)a(n+1)+1/3,a(n+1) = s(n+1)-s(n)= (2/3)a(n+1) - (2/3)a(n),a(n+1) = -2a(n),{a(n)}是首项为1,公比为-2的等比数列.a(n) = (-2)^(n-1)...