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题目：已知函数f(x)=x/(3x+1)数列an满足a1=1 a(n+1)=f(an)
（1）求数列的通项公式
（2）Sn=a1*a2+a2*a3+……+an*a(n+1)求Sn并算极限
解：
a(n+1)=f(an)=an/(3an+1)
两边取倒数
1/a(n+1)=3+1/an
1/a(n+1)-1/an=3
数列{1/an}是以1为首项，公差为3的等差数列
通项为 1/an=3(n-1)+1=3n-2
所以数列{an}的通项为 an=1/(3n-2)
(2)
Sn
=a1*a2+a2*a3+……+an*a(n+1)
=(1/1)*(1/4)+(1/4)*(1/7)+……+[1/(3n-2)]*[1/(3n+1)]
=1/(1*4)+1/(4*7)+……+1/[(3n-2)*(3n+1)]
=[1/1-1/4]/3+[1/4-1/7]/3+……+[1/(3n-2)-1/(3n+1)]/3
=[1-1/4+1/4-1/7+……+1/(3n-2)-1/(3n+1)]/3
=[1-1/(3n+1)]/3
=n/(3n+1)
--lim Sn=lim n/(3n+1)=1/3
n→+∞