在各项均为正数的数列an中,前n项与sn满足sn=1/8（an+2）²（1）求证an为等差数列

(1)Sn=1/8(an+2)^2,S(n-1)=1/8[a(n-1)+2]^2,an=Sn-S(n-1)=1/8[an^2+4an-a(n-1)^2-4a(n-1)] 8an=an^2+4an-a(n-1)^2-4a(n-1) an^2-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-...