a1=1,a2=3/2,a3=11/6,a4=43/22求通项公式

a(n+1)=2-1/2(2/an-1)=5/2-1/an
a(n+1)-2=1/2-1/an=(an-2)/2an
所以1/(a(n+1)-2)=2an/(an-2)=2+4/(an-2)
令bn=1/(an-2)
则b(n+1)=4bn+2
b(n+1)+2/3=4(bn+2/3)
即数列{bn+2/3}为首项b1+2/3=-1/3,公比为4的等比数列
bn+2/3=-1/3*4^(n-1),bn=-1/3[4^(n-1)+2]
所以1/(an-2)=-1/3[4^(n-1)+2]
即an=2-3/[4^(n-1)+2]