求问数学题.需要详解已知函数f（x）x/3x+1,数列{a n}满足a1=1,a n+1=f(a n)(n∈N*)『1』

1.a n+1=f(an)即a n+1=an/(3an+1)
化解得an=3an*a n+1+a n+1
1/a n+1 -1/an =3
a1=1,a2=1/4
1/a n -1/a n-1 +1/a n-1 -1/a n-2 .1/a2 -1/a1=((1/a2 -1/a1)*3(n-1)+(1/a2 -1/a1))(n-1)/2
1/a n-1/a1=(9n^2-15n+6)/2
1/a n -1=(9n^2-15n+6)/2
an=2/(9n^2-15n+8)
2.由上可知,3an*a n+1=a n+1-an
sn=a1a2+a2a3+…+an an+1
=1/3(a2-a1+a3-a2+...+a n+1-an)
=1/3(an+1-a1)
=1/3(2/(9(n+1)^2-15(n+1)+8)-1)
=1/3(2/(9n^2+3n+2)-1)
=-(3n^2+n)/(9n^2+3n+2)