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)²-[a(n-1)+2]²}
=(1/8)[(an+a(n-1)+4][an-a(n-1)]
8an=an²-a(n-1)²+4an-4a(n-1)
[an+a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0
两边同除以an+a(n-1)
an-a(n-1)=4
所以{an}是公差为4的等差数列