题目

已知函数f(x)=x/(3x+1),数列｛an｝满足a1=1,an+1=f(an)(n∈N*）
记Sn=a1a2+a2a3+...+anan+1,求Sn

提问时间：2020-12-09

答案

由题意
a(n+1)=an/(3an+1)
取倒数得
[1/a(n+1)]=[1/an]+3
故数列{1/an}是公差为3首项为1/a1=1的等差数列
故1/an=1+3(n-1)
=3n-2
∴
sn=1/(3n-2)
∴
Sn=1/(1*4)+1/(4*7)...+1/(3n-2)(3n+1)
=(1/3)[1-(1/4)+(1/4).+(1/(3n-2))-1/(3n+1)]
=(1/3)3n/(3n+1)
=n/(3n+1)