an=[7a(n-1)-3]/[3a(n-1)+1],
an-1=[7a(n-1)-3]/[3a(n-1)+1]-1,
an-1=[4a(n-1)-4]/[3a(n-1)+1],
取倒數(shù)得：
1/ (an-1)= [3a(n-1)+1]/ [4a(n-1)-4],
1/ (an-1)= [3a(n-1)-3+4]/ [4a(n-1)-4],
1/ (an-1)=3/4+4/ [4a(n-1)-4],
1/ (an-1)=3/4+1/ [a(n-1)-1],
這說明數(shù)列{1/ (an-1)}是等差數(shù)列,首項為1/(a1-1)=1,公差為3/4.
1/ (an-1)=1+(n-1)•3/4,
1/ (an-1)=(3n+1)/4,
an-1=4/(3n+1),
an=(3n+5) /(3n+1).