已知数列[an]满足Sn=0.25an+1,求a1+a3+a5+……+a2n-1的极限

Sn=0.25an+1,
n=1时,a1=0.25a1+1,a1=4/3
n>1时,S（n-1）=0.25a（n-1）+1
两式相减,得an=0.25an-0.25a(n-1),an/a(n-1)=-1/3
an=4/3•(-1/3)^(n-1)
a1+a3+a5+……+a2n-1=4/3+4/3•1/9+4/3•1/81+……4/3•(1/9)^(n-1)
a1+a3+a5+……+a2n-1的极限=(4/3)/(1-1/9)=3/2