高一函数数列综合问题(急~)求证:1+f(1/5)+f(1/11)+...+f(1/(n^2+3n+1))=-f(1/(n+2))已知函数f(x)在(-1,1)上有定义,f(1/2)=-1,满足:x,y∈(-1,1)时,有f(x)+f(y)=f((x+y)/(1+xy)).(1)证明在(-1,1)上恒有f(-x)=-f(x
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![高一函数数列综合问题(急~)求证:1+f(1/5)+f(1/11)+...+f(1/(n^2+3n+1))=-f(1/(n+2))已知函数f(x)在(-1,1)上有定义,f(1/2)=-1,满足:x,y∈(-1,1)时,有f(x)+f(y)=f((x+y)/(1+xy)).(1)证明在(-1,1)上恒有f(-x)=-f(x](/uploads/image/z/3274568-8-8.jpg?t=%E9%AB%98%E4%B8%80%E5%87%BD%E6%95%B0%E6%95%B0%E5%88%97%E7%BB%BC%E5%90%88%E9%97%AE%E9%A2%98%EF%BC%88%E6%80%A5%7E%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A1%2Bf%281%2F5%29%2Bf%281%2F11%29%2B...%2Bf%281%2F%28n%5E2%2B3n%2B1%29%29%3D-f%281%2F%28n%2B2%29%29%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%EF%BC%88-1%2C1%EF%BC%89%E4%B8%8A%E6%9C%89%E5%AE%9A%E4%B9%89%2Cf%281%2F2%29%3D-1%2C%E6%BB%A1%E8%B6%B3%EF%BC%9Ax%2Cy%E2%88%88%EF%BC%88-1%2C1%EF%BC%89%E6%97%B6%2C%E6%9C%89f%28x%29%2Bf%28y%29%3Df%28%28x%2By%29%2F%281%2Bxy%29%29.%EF%BC%881%EF%BC%89%E8%AF%81%E6%98%8E%E5%9C%A8%28-1%2C1%EF%BC%89%E4%B8%8A%E6%81%92%E6%9C%89f%28-x%29%3D-f%28x)
高一函数数列综合问题(急~)求证:1+f(1/5)+f(1/11)+...+f(1/(n^2+3n+1))=-f(1/(n+2))已知函数f(x)在(-1,1)上有定义,f(1/2)=-1,满足:x,y∈(-1,1)时,有f(x)+f(y)=f((x+y)/(1+xy)).(1)证明在(-1,1)上恒有f(-x)=-f(x
高一函数数列综合问题(急~)求证:1+f(1/5)+f(1/11)+...+f(1/(n^2+3n+1))=-f(1/(n+2))
已知函数f(x)在(-1,1)上有定义,f(1/2)=-1,满足:x,y∈(-1,1)时,有f(x)+f(y)=f((x+y)/(1+xy)).
(1)证明在(-1,1)上恒有f(-x)=-f(x);
(2)数列{an}满足a1=1/2,a(n+1)=2an/(1+an^2),设xn=f(an),求{xn}的通项;
(3)求证:1+f(1/5)+f(1/11)+...+f(1/(n^2+3n+1))=-f(1/(n+2))
第(1)(2)的答案已经做出来了,这里只要第(3)问的详解,急
注:{xn}的通项已经算出,为-2^(n-1)
高一函数数列综合问题(急~)求证:1+f(1/5)+f(1/11)+...+f(1/(n^2+3n+1))=-f(1/(n+2))已知函数f(x)在(-1,1)上有定义,f(1/2)=-1,满足:x,y∈(-1,1)时,有f(x)+f(y)=f((x+y)/(1+xy)).(1)证明在(-1,1)上恒有f(-x)=-f(x
(1)令x=y=0,则有f(0)+f(0)=f(0).所以f(0)=0,
再令y=-x,则有f(x)+f(-x)=f((x-x)/(1-x^2))=f(0)=0.
所以f(-x)=-f(x);
(2)a1>0,所以a(n+1)=2an/(1+an^2)>0,又2an
俺要上课了,建议LZ用数归法试一下。我用了该法,不难的。