a(n+1)=(5a(n)+6)/(a(n)+4),a(n+1)+2=(5a(n)+6)/(a(n)+4)+2,a(n+1)+2=(7a(n)+14)/(a(n)+4)取倒数得:1/[a(n+1)+2]=(a(n)+4)/ [7(a(n)+2)],1/[a(n+1)+2]=(a(n)+2+2)/ [7(a(n)+2)],1/[a(n+1)+2]=1/7+2/ [7(a(n)+2)],设1/[a(n)+2]=bn,b1=1/7.则b(n+1)= 1/7+2/ 7*bn,b(n+1)-1/5=2/ 7*(bn-1/5),所以{bn-1/5}是等比数列,公比是2/7,首项是b1-1/5=-2/35.bn-1/5=-... |