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已知函数f(x)=x^2/1+x^2 那么f(1)+f(2)+f(3)+...+f(2009)+f(1/2)+f(1/3)+...+f(1/2009)=?
题目详情
已知函数f(x)=x^2/1+x^2 那么f(1)+f(2)+f(3)+...+f(2009)+f(1/2)+f(1/3)+...+f(1/2009)=?
▼优质解答
答案和解析
f(x)=x^2/(1+x^2)
f(n)+f(1/n)
=n^2/(1+n^2)+(1/n)^2/[1+(1/n)^2]
=n^2/(1+n^2)+1/(1+n^2)
=(n^2+1)/(n^2+1)
=1
f(1)+f(2)+f(3)+...+f(2009)+f(1/2)+f(1/3)+...+f(1/2009)
=f(1)+[f(2)+f(1/2)]+[f(3)+f(1/3)]+.+[f(2009)+f(1/2009)]
=1/2+1+1+.+1
=2008又1/2
f(n)+f(1/n)
=n^2/(1+n^2)+(1/n)^2/[1+(1/n)^2]
=n^2/(1+n^2)+1/(1+n^2)
=(n^2+1)/(n^2+1)
=1
f(1)+f(2)+f(3)+...+f(2009)+f(1/2)+f(1/3)+...+f(1/2009)
=f(1)+[f(2)+f(1/2)]+[f(3)+f(1/3)]+.+[f(2009)+f(1/2009)]
=1/2+1+1+.+1
=2008又1/2
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