题目内容
For 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 f(x)=(x+1)/(x-1),then[f(2)]+[f(3)]+…+[f(100)]=分析:利用函数f(x)=
,可得出f(2)…f(100)代表的数据,从而得出[f(2)]=3,[f(3)]=2,[f(4)]=[f(5)]=…[f(100)]=1,的值,进而求出结果.
| x+1 |
| x-1 |
解答:解:∵f(x)=
,
∴f(2)=
=3,f(3)=
=2,
f(4)=
=
,f(5)=
=
,…f(100)=
=
,
∴[f(2)]=3,[f(3)]=2,[f(4)]=[f(5)]=…[f(100)]=1,
∴[f(2)]+[f(3)]+…+[f(100)],
=3+2+1+…+1,
=5+1×97,
=102.
故答案为:102.
| x+1 |
| x-1 |
∴f(2)=
| 2+1 |
| 2-1 |
| 3+1 |
| 3-1 |
f(4)=
| 4+1 |
| 4-1 |
| 5 |
| 3 |
| 5+1 |
| 5-1 |
| 3 |
| 2 |
| 100+1 |
| 100-1 |
| 101 |
| 99 |
∴[f(2)]=3,[f(3)]=2,[f(4)]=[f(5)]=…[f(100)]=1,
∴[f(2)]+[f(3)]+…+[f(100)],
=3+2+1+…+1,
=5+1×97,
=102.
故答案为:102.
点评:此题主要考查了取整函数的性质,以及由已知得出[f(2)]…[f(100)]代表的数据,这是解决问题的关键.
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