a(n+1)=1/3 sn =>s(n+1)-sn=1/3 sn => s(n+1)=4/3*sn s1=a1=1 => sn=(4/3)^(n-1) 对任意自然数n>=1 => a(n+1)=s(n+1)-sn=(4/3)^(n-1)/3 对任意自然数n>=1 所以 a1=1 an=(4/3)^(n-2)/3 (n>=2)
