已知函数f(x)=x/(3X+1),数列an满足a1=1,a(n+1)=f(an)(n为正整数)(1)求数列an的通项公

f(x)=x/(3X+1)
→1/f(x)=(3x+1)/x
→1/f(x)=3+1/x
即1/a(n+1)=1/an+3
故1/an=1/a(n-1)+3=1/a(n-2)+3*2
=1/a(n-3)+3*3
=……
=1/a1+3*(n-1)
=1/1+3(n-1)
=3n-2
故an=1/(3n-2) (n≥2)
又n=1时,1=a1=1/(3-2)亦满足
故an=1/(3n-2)