（1）b=[a+p]/[a-p]
=(an+p^2/an+2p)/(an+p^2/an-2p)
=[(an+p)/(an-p)]^2=bn^2,
b1=(a1+p)/(a1-p)=3,
lg[b]=2lg(bn),
∴lgbn=lgb1*2^(n-1)=lg3*2^(n-1),
∴bn=10^[lg3*2^(n-1)]=3^[2^(n-1)].
(2)由(1),(an+p)/(an-p)=3^[2^(n-1)],
∴an+p=(an-p)*3^[2^(n-1)],
∴p{1+3^[2^(n-1)]={3^[2^(n-1)-1]an,
∴an=p{3^[2^(n-1)+1}/{3^[2^(n-1)]-1},
∴an-p=2p/{3^[2^(n-1)]-1},
∴（an-p)/[a-p]={3^(2^n)-1}/{3^[2^(n-1)-1}=3^[2^(n-1)]+1.
(3)a2=5p/4,a3=41p/40,p>0,
S2=a1+a2=(2+5/4)p=3时由（2）[a-p]/(an-p)