n-b(n-1)=1/[2-4/(an-1)]-1/[a(n-1)-2]
=a(n-1)/[2a(n-1)-4]-2/[2a(n-1)-4]
=[a(n-1)-2]/[2a(n-1)-4]
=1/2
所以数列{bn}是以b1=1/2为首项,公差为1/2的等差数列.
所以bn=n/2,故an=2+2/n
n-b(n-1)=1/[2-4/(an-1)]-1/[a(n-1)-2]
=a(n-1)/[2a(n-1)-4]-2/[2a(n-1)-4]
=[a(n-1)-2]/[2a(n-1)-4]
=1/2
所以数列{bn}是以b1=1/2为首项,公差为1/2的等差数列.
所以bn=n/2,故an=2+2/n