急！！一道高中数学题（最好有过程
数列{an}的前n项和Sn满足an=5Sn-3(n∈N*)
n=1: a1=5S1 -3 = 5*a1 -3 ===> a1=S1=3/4
又：an=5Sn-3 ==> Sn -S(n-1)=5Sn -3
==> (Sn -3/5) = (-1/4)[S(n-1) -3/5]
==> 数列{Sn-3/5}为公比q=-1/4、首项=3/4-3/5=3/20的等比数列
==> Sn -3/5 = (3/20)(-1/4)^(n-1)
a(2n-1)=S(2n-1)-S(2n-2)=12(1/4)^(2n)
a(2n-3)=S(2n-3)-S(2n-4)=12(1/4)^(2n-2)
==> a...全部
数列{an}的前n项和Sn满足an=5Sn-3(n∈N*)
n=1: a1=5S1 -3 = 5*a1 -3 ===> a1=S1=3/4
又：an=5Sn-3 ==> Sn -S(n-1)=5Sn -3
==> (Sn -3/5) = (-1/4)[S(n-1) -3/5]
==> 数列{Sn-3/5}为公比q=-1/4、首项=3/4-3/5=3/20的等比数列
==> Sn -3/5 = (3/20)(-1/4)^(n-1)
a(2n-1)=S(2n-1)-S(2n-2)=12(1/4)^(2n)
a(2n-3)=S(2n-3)-S(2n-4)=12(1/4)^(2n-2)
==> a1=3/4, a(2n-1)/a(2n-3) = 1/16
==> {a1+a3+…a(2n-1)。
。。}为首项=3/4、公比=1/16的等比数列
因此，lim[a1+a3+……a(2n-1)]
= (3/4)/[1- (1/16)]
= 4/5
。收起
