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
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 16:13:37
![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](/uploads/image/z/6913141-61-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx%2F2x%2B1%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1a%28n%2B1%29%3Df%28an%29%E8%AF%81%E6%98%8E%E6%98%AF1%2Fan%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+%E8%AE%B0sn%3D2%2Fa1%2B2%2A2%2Fa2%2B2%2A3%2Fa3%2B.%2B2%2An%2Fan%E6%B1%82sn)
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