题目内容
新运算f(1)=1+
,f(2)=1+
,f(3)=1+
,以此类推.计算:f(1)×f(2)×…×f(100)= .
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考点:规律型:数字的变化类
专题:
分析:由f(1)=1+
=
,f(2)=1+
=
,f(3)=1+
=
,…f(100)=1+
=
,直接代入交错约分得出答案即可.
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| 100 |
| 102 |
| 100 |
解答:解:∵f(1)=1+
=
,f(2)=1+
=
,f(3)=1+
=
,…f(100)=1+
=
,
∴f(1)×f(2)×…×f(100)
=
×
×
×
×…×
×
=
=5151.
故答案为:5151.
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| 100 |
| 102 |
| 100 |
∴f(1)×f(2)×…×f(100)
=
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| 101 |
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=
| 101×102 |
| 1×2 |
=5151.
故答案为:5151.
点评:此题考查数字的变化规律,找出数字之间的运算规律,利用规律解决问题.
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