Sn=（1/4an+1）,求lim(a2+a4+a6+...+a2n)
Sn=（1/4an+1）照写
S（n+1）= 0.25a(n+1)+1
下减去上
a(n+1)=0.25a(n+1)-0.25an
--- a(n+1)= 1/3*an
显然 等比数列
a1=S1=0.25a1+1---- a1=4/3
an= a1*(1/3)^(n-1)=4*(1/3)^n
a2n=4/9*(1/9)^(n-1)=4*(1/9)^n
S2n= 4/9*(1-(1-1/9)^n)/(1-1/9)
lim(a2+a4+a6+...+a2n)
= 4/9/(1-1/9)
=1/2=0.5