(一）（1）由a1=1,S(n+1)=4an+2.可得：a1=1,a2=5,a3=16.a4=44.∴由bn=(an)/2^n得：b1=2/4,b2=5/4,b3=8/4,b4=11/4.顯然,b1,b2,b3,b4是等差數(shù)列.（2）S(n+1)=4an+2,===>Sn=4a(n-1)+2.兩式相減得a(n+1)=4an-4a(n-1).===>a(n+1)-2an=2[an-2a(n-1)].∴a(n+1)-2an=3×2^(n-1)=(3/4)×2^(n+1).===>[a(n+1)/2^(n+1)]-[an/2^n]=3/4.===>b(n+1)-bn=3/4.∴{bn}是等差數(shù)列.其通項bn=(3n-1)/4.===>an=(3n-1)×2^(n-2),(n=1,2,3,...).（二）an=1/[√(n+1)+√n]=√(n+1)-√n.∴Sn=√(n+1)-1.S100=√101-1.(2)1²-2²+3²-4²+5²-6²+...+99²-100²=-[(2²-1²)+(4²-3²)+(6²-5²)+...+(100²-99²)]=-[1+2+3+...+100]=-5050