已知数列{an}满足Sn=(1/4)an+1,求lim(a1+a3+a5+...+a2n-1)的值

Sn=an/4+1,
——》S(n-1)=a(n-1)/4+1,
——》an=Sn-S(n-1)=an/4+1-a(n-1)/4-1=an/4-a(n-1)/4,
——》an/a(n-1)=-1/3=q,
S1=a1=a1/4+1,
——》a1=4/3,
——》an=a1*q^(n-1),
设数列{bn}为bn=a(2n-1),则：
b1=a1,bn=a(2n-1)=a1*q^(2n-1-1)=b1*(q^2)^(n-1),
即数列{bn}为首项b1=a1=4/3,公比q‘=q^2的等比数列,
其前n项和
Tn=a1+a3+a5+...+a2n-1
=b1*(1-q'^n)/(1-q')
=a1*(1-q^2n)/(1-q^2)
=4/3*[1-(-1/3)^2n]/[1-(-1/3)^2]
=3/2[1-(1/9)^n],
limn→∞Tn
=limn→∞3/2[1-(1/9)^n]
=3/2(1-0)
=3/2.

用求和公式，讲带q一项视为0