俺来看看
a(n+1)=1/(2-an)
a(n+1) -1 = 1/(2-an) -1
∴a(n+1) -1 =(an-1)/(2-an)
取倒数则1/[a(n+1) -1]=(2-an)/(an-1),即1/[a(n+1) -1]=1/(an -1) -1
∴1/(an -1)是以-1为公差的等差数列,
1/(a1 -1)=-1
∴1/(an -1)=-n
∴an=(n-1)/n