已知函数f(x)=x/2x+1数列an满足a1=1a(n+1)=f(an)证明是1/an等差数列 记sn=2/a1+2*2/a2+2*3/a3+.+2*n/an求sn

已知函数f(x)=x/2x+1数列an满足a1=1a(n+1)=f(an)证明是1/an等差数列 记sn=2/a1+2*2/a2+2*3/a3+.+2*n/an求sn
已知函数f(x)=x/2x+1数列an满足a1=1a(n+1)=f(an)
证明是1/an等差数列 记sn=2/a1+2*2/a2+2*3/a3+.+2*n/an求sn

已知函数f(x)=x/2x+1数列an满足a1=1a(n+1)=f(an)证明是1/an等差数列 记sn=2/a1+2*2/a2+2*3/a3+.+2*n/an求sn
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-1
2n×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)＝（4n-1)n(n+1)/3