2005全国"希望杯"数学竞赛试题[一道英语数学题]For-a-real-number-a,let[a]denote-the maximum-integer-which-does-not-exceed-a.For example,[3.1]=3,[-1.5]=-2,[0.7]=0.Now let.x+1f(x)=—.x-1,then[f(2)]+[f(3)]+L+[f(100)](real-number:实数,t
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2005全国"希望杯"数学竞赛试题[一道英语数学题]For-a-real-number-a,let[a]denote-the maximum-integer-which-does-not-exceed-a.For example,[3.1]=3,[-1.5]=-2,[0.7]=0.Now let.x+1f(x)=—.x-1,then[f(2)]+[f(3)]+L+[f(100)](real-number:实数,t
2005全国"希望杯"数学竞赛试题[一道英语数学题]
For-a-real-number-a,let[a]denote-the maximum-integer-which-does-not-exceed-a.For example,[3.1]=3,[-1.5]=-2,[0.7]=0.Now let
.x+1
f(x)=—
.x-1,then[f(2)]+[f(3)]+L+[f(100)]
(real-number:实数,the maximum-integer-which-does-not-exceed-a:不超过a的最大整数)
2005全国"希望杯"数学竞赛试题[一道英语数学题]For-a-real-number-a,let[a]denote-the maximum-integer-which-does-not-exceed-a.For example,[3.1]=3,[-1.5]=-2,[0.7]=0.Now let.x+1f(x)=—.x-1,then[f(2)]+[f(3)]+L+[f(100)](real-number:实数,t
.x+1
f(x)=— = 1+[2/(x-1)]
.x-1
f(2)=1+2=3;
f(3)=1+2/2=2;
f(4)=1+2/3=4;
[f(2)]=3;
[f(3)]=2;
[f(2)]+[f(3)]+L+[f(100)]
=3+2+(100-2)
=103