设数列{an}是首项为1的等比数列,若{1/[2an+a(n+1)]}是等差数列,则[1/(2a1)+1/(a2)]+[1/(2a2)+1/(a3)]+[1/(2a3)+1/(a4)]+……+[1/(2a2012)+1/(a2013)]的值

问题描述:

设数列{an}是首项为1的等比数列,若{1/[2an+a(n+1)]}是等差数列,则[1/(2a1)+1/(a2)]+[1/(2a2)+1/(a3)]+[1/(2a3)+1/(a4)]+……+[1/(2a2012)+1/(a2013)]的值等于_______

由等比设A1=1,A2=Q,A3=Q方
由等差得1/(2+Q)+1/(2Q方+Q立)=2/(2Q+Q方)解得Q=1
即公比为1,公差为0
原式=2012/3