an=a(n-1)/[3a(n-1)+1],取倒数.(n≥2)
1/an=3+1/a(n-1),1/a2=4.
即n≥2时,{1/an}是第二项a2=4,公差d=3的等差数列.
1/an=1/a2+(n-2)d=3n-2.(n≥2),又n=1也满足.
∴1/an=3n-2,an=1/(3n-2).
bn=ana(n+1)=1/(3n-2)(3n+1)=1/3(1/(3n-2)-1/(3n+1))
Sn=1/3(1/(3-2)-1/(3+1)+1/(6-2)-1/(6+1)+1/(9-2)-1/.-1/(3n+1))
=1/3(1-1/(3n+1).