已知正数数列an的前n项和为Sn,若an和2的等差中项与Sn和2的等比中项相等

1、
由题意,得
(a1+2)/2=√(2a1)
整理,得
(a1-2)²=0
a1-2=0 a1=2
(an+2)/2=√(2Sn)
整理,得
8Sn=(an+2)²
8Sn-1=[a(n-1)+2]²
8an=(an+2)²-[a(n-1)+2]²=an²+4an-a(n-1)²-4a(n-1)
an²-a(n-1)²-4an-4a(n-1)=0
[an+(a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-4]=0
数列是正数数列,an+a(n-1)>0,要等式成立,只有an-a(n-1)-4=0
an-a(n-1)=4,为定值.
数列{an}是以2为首项,4为公差的等差数列.
an=2+4(n-1)=4n-2
数列{an}的通项公式为an=4n-2
2、
bn=0.5[a(n+1)/an+an/a(n+1)]
=[(4n+2)/(4n-2)+(4n-2)/(4n+2)]/2
=[1+4/(4n-2)+1-4/(4n+2)]/2
=1+2/[(2n-1)(2n+1)]
=1+1/(2n-1)-1/(2n+1)
b1+b2+...+bn
=n+1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)
=n+1-1/(2n+1)