(i)Sn=(√a1 -√a2)/(a1-a2)+(√a2 -√a3)/(a2-a3)+...+(√an -√a(n+1))/(an-a(n+1))=(-1/2)(√a1 -√a2+√a2 -√a3+...+√an -√a(n+1))=-(1/2)(√a1 -√a(n+1)),a100=25+100*2=15^2,得S100=5
（ii）令n=1,知p=1,当n>=2时,Sn-S(n-1)=[pn/√a1+√a(n+1)]-(n-1)*p/[√a1+√an]=1/[√an +√a(n+1)]化为n/[√a1+√a(n+1)]-n/[√a1+√an]=1/[√an +√a(n+1)]-1/[√a1+√an],去分母,化简,得n[an-a(n+1)]=a1-a(n+1),即a(n+1)/n+a1/n(n-1)=an/(n-1),即 a(n+1)/n-a1/n=an/(n-1)-a1/(n-1)=.=a2-a1=d为定值,所以a(n+1)=a1+nd,an=a1+(n-1)d,则知{an}为等差数列