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)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)+……+(1/2a2012+1/a2013)=?

Question

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)
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)+……+(1/2a2012+1/a2013)=?

Answer 1

设an=q^(n-1),{bn}={1/[2an+a(n+1)] }是等差数列,所以有2bn=bn-1+bn+1,得到q=1,所以an=a1=1,所以原式=3018