a(n+1)=2a(n)/[4-a(n)],若a(n+1)=0,则a(n)=0,...,a(1) = 0,与a(1)=1矛盾,因此,a(n)不为0.1/a(n+1) = [4-a(n)]/[2a(n)] = 2/a(n) - 1/2,1/a(n+1) - 1/2 = 2/a(n) - 2/2 = 2[1/a(n) - 1/2],{1/a(n) - 1/2}是首项为1/a(...