b(n+1)=1/(2-bn)
b(n+1) -1=(1-2+bn)/(2-bn)=(bn-1)/(2-bn)
1/[b(n+1)-1]=(2-bn)/(bn-1)=1/(bn -1) -1
1/[b(n+1)-1]-1/(bn -1)=-1,为定值.
1/(b1-1)=1/(3/4-1)=-4,数列1/(bn -1)是以-4为首项,-1为公差的等差数列.
1/(bn -1)=-4+(-1)(n-1)=-n-3
bn=-1/(n+3) +1=(n+2)/(n+3)
b1=(1+2)/(1+3)=3/4,同样满足通项公式
数列{an}的通项公式为an=(n+2)/(n+3).