∵Sn=1-(2/3)an
∴S1=a1=1-(2/3)a1
a1=3/5
同理,有：S2=a2+a1=1-(2/3)a2
解之得：a2=6/25
S3=a1+a2+a3=1-(2/3)a3
a3=12/125
∴S(n+1)=1-(2/3)a(n+1)
又S(n+1)-Sn=a(n+1)
∴a(n+1)=[1-(2/3)a(n+1)]-[1-(2/3)an]
a(n+1)=(2/5)an
∴an=(2/5)a(n-1）
又a1=3/5
a2=(3*2^1)/5^2
a3=(3*2^2)/5^3
……
∴an=[3*2^(n-1)]/5^n