设an是正数组成的数列,其前n项和为Sn,并且对于所有的自然数n,an与2的等差中项等于Sn与2的等比中项．

(an +2)/2=√(2Sn)
8Sn=(an+2)²
n=1时,8S1=8a1=(a1+2)²
(a1-2)²=0
a1=2
n≥2时,8Sn=(an +2)² 8S(n-1)=[a(n-1)+2]²
8Sn-8S(n-1)=8an=(an+2)²-[a(n-1)+2]²
(an -2)²=[a(n-1)+2]²
an -2=a(n-1)+2或an-2=-a(n-1)-2
an-a(n-1)=4或an=-a(n-1)(数列各项均为正,舍去)
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 你求的通项公式是对的.
bn=(1/2)[a(n+1)/an +an/a(n+1)]
=(1/2)[(4n+2)/(4n-2)+(4n-2)/(4n+2)]
=(1/2)[(2n+1)/(2n-1)+(2n-1)/(2n+1)]
=(1/2)[(2n+1)²+(2n-1)²]/[(2n-1)(2n+1)]
=(4n²+1)/(4n²-1)
=(4n²-1+2)/(4n²-1)
=1 +2/[(2n-1)(2n+1)]
=1 +1/(2n-1) -1/(2n+1)
b1+b2+...+bn
=1+1/1-1/3+1+1/3-1/5+...+1+1/(2n-1)-1/(2n+1)
=n+1-1/(2n+1)