各项均为正数的等比数列{an}中,前n项和为Sn=（2分之an+1）∧2 （1）求数列{an}通项公式 （2）若a1a2

(1)
Sn = [(an +1)/2]^2
n=1
4a1=(a1+1)^2
a1^2-2a1+1=0
a1=1
an=Sn -S(n-1)
4an = (an+1)^2-(a(n-1) +1)^2
(an)^2-[a(n-1)]^2 - 2[an + a(n-1)]=0
[an+a(n-1)].[an-a(n-1)-2]=0
an-a(n-1)-2=0
an-a1=2(n-1)
an=2n-1
(1)
1/[an.a(n+1)] = 1/[(2n-1)(2n+1)]
= (1/2)[1/(2n-1) - 1/(2n+1)]
1/(a1.a2)+1/(a2.a3)+...+1/[an.a(n+1)]
=(1/2)[1- 1/(2n+1)]
k≥1/2