a2=(1/3)S1=(1/3)a1=1/3
a3=(1/3)S2=(1/3)(a1+a2)=4/9
a4=(1/3)S3=(1/3)(a1+a2+a3)=16/27
a(n+1)=(1/3)Sn
a(n+2)=(1/3)S(n+1)
a(n+2)-a(n+1)=(1/3)[S(n+1)-Sn]=(1/3)a(n+1)
a(n+2)/a(n+1)=4/3
所以:{an}是公比为4/3的等比数列
an=a1*q^(n-1)=(4/3)^(n-1)
{a2n}是公比为(4/3)^2=16/9的等比数列,首项为a2=4/9
a2+a4+a6+……+a2n
=(4/9)[(16/9)^n-1}/[(16/9)-1]
=(4/7)[(16/9)^n-1]