已知等比数列{an}的公比q=-(1/3),则极限(a2+a4+...+a2n)/(a1+a2+...+an)=
问题描述：
已知等比数列{an}的公比q=-(1/3),则极限(a2+a4+...+a2n)/(a1+a2+...+an)=
答
an = a1.(-1/3)^(n-1)
a2+a4+...+a2n = a1[ (1/3)^2+(1/3)^4+...+(1/3)^(2n) ]
= (a1/8) [ 1 - 1/3^(2n)]
a1+a2+...+an = (3a1/4) [ 1 - (-1/3)^n ]
lim(n->∞) (a2+a4+...+a2n)/(a1+a2+...+an)
=lim(n->∞) (1/6)[ 1 - 1/3^(2n)] /[ 1 - (-1/3)^n ]
=1/6为什么我的老师的答案是1/2啊不好意思an = a1.(-1/3)^(n-1)a2+a4+...+a2n = -a1[ (1/3)+(1/3)^3+...+(1/3)^(2n-1) ]= -(3a1/8) [ 1 - 1/3^(2n)]a1+a2+...+an = (4a1/3) [ 1 - (-1/3)^n ]lim(n->∞) (a2+a4+...+a2n)/(a1+a2+...+an)=lim(n->∞) (-1/2)[ 1 - 1/3^(2n)] /[ 1 - (-1/3)^n ]=-1/2