正数列{An}An与2的等差中项等于Sn与2的等比中项.1.求{An}是等差数列.2.求An的通项公式.

1.
(an + 2)/2=√(2Sn)
an + 2=2√(2Sn)
(an + 2)²=8Sn…………①
∴[a(n-1) + 2]²=8S(n-1)…………②
①－②,得
(an + 2)²－[a(n-1) + 2]²=8Sn－8S(n-1)
an² + 4an + 4 - a(n-1)² - 4a(n-1) - 4=8an
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) -4]·[an + a(n-1)]=0
∵an＞0,∴an + a(n-1)≠0
∴an - a(n-1) -4=0,即an - a(n-1) = 4
∴数列{an}是等差数列.
2.
(a1+2)/2=√(2S1)=√(2a1)
∴a1=2
∴an=a1+(n-1)d=2+4(n-1)=4n-2
3.
Bn=1/2[(An+1/An)+(An/An+1)]
=1/2[(4n+2)/(4n-2)+(4n-2)/(4n+2)]
=1/2[(4n-2+4)/(4n-2) + (4n+2-4)/(4n+2)]
=1/2[1 + 4/(4n-2) + 1 - 4/(4n+2)]
=1/2[2 + 4/(4n-2) - 4/(4n+2)]
=1 + 2/(4n-2) - 2/(4n+2)
=1 + 1/(2n-1) - 1/(2n+1)
∴Sbn= [1 + (1- 1/3)] + [1+ (1/3 - 1/5)] + [ 1 + (1/5 - 1/7)]+ ……+[1 + 1/(2n-1) - 1/(2n+1)]
= n + [1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-1) - 1/(2n+1)]
= n + [1 - 1/(2n+1)]
=(2n²+3n)/(2n+1)