∵a(n+1)=1/(2-an)
∴1/[a(n+1)-1]=1/[1/(2-an)-1]
∵1/[a(n+1)-1]=1/(an-1)-1
∴1/[an-1]为等差数列
∴1/(an-1)
=-1*(n-1)+1/(a1-1)
=-(n-1)+1/(-1/2)
=-n-1
∴an=1-1/(n+1)