已知函数f(x)=x/2x+1数列an满足a1=1a(n+1)=f(an)

an+1=an/(2an + 1) 1/an+1=(2an + 1)/2=1/an +2 1/an+1-1/an=2所以1/an是首项为1,公差为2的等差,1/an=1+2(n-1)=2n-12n×1/an=2n(2n-1)=4n^2-2n所以sn=4(1^2+2^2+...+n^2)-2(1+2+...n)＝4×n(n+1)(2n+1)/6 －n(n+1)＝...