数列题,已知函数f(x)=x/(3x+1),数列{an}满足a1=1,a(n+1)=f(an),n是正整数求{an}的通

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数列题,
已知函数f(x)=x/(3x+1),数列{an}满足a1=1,a(n+1)=f(an),n是正整数
求{an}的通项公式
记Sn=a1*a2+a2*a3+a3*a4+…+an*a(n+1),求Sn

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)Sn=a1*a2+a2*a3+a3*a4+…+an*a(n+1)Sn=1/1*4+1/4*7+1/7*10...+1/(3n-2)(3n+1)=1/3(1-1/4+1/4-1/7+1/7-1/10+...+1/(3n-2)-1/(3n+1))=1/3(1-1/(3n+1))=1/3(3n/(3n+1))=n/(3n+1)

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