题目内容
①1993+1994+1995+1996+1997+1998+1999+2000=
②(0.75×42.7+57.3-0.573×25)÷3×7972=
解:(1)1993+1994+1995+1996+1997+1998+1999+2000,
=(1993+2000)×8÷2,
=3993×4,
=15972.
(2)(0.75×42.7+57.3-0.573×25)÷3×7972,
=(0.75×42.7+57.3-57.3×0.25)÷3×7972,
=[0.75×42.7+57.3×(1-0.25)]÷3×7972,
=[0.75×42.7+57.3×0.75]÷3×7972,
=[0.75×(42.7+57.3)]÷3×7972,
=0.75×100÷3×7972,
=75÷3×7972,
=25×7972,
=25×(4×1993),
=25×4×1993,
=100×1993,
=199300.
分析:(1)这是一道典型的等差数列求和.首项与末项之和×项数÷2,
首项=这个数列的第一个数1993.末项=最后一个数2000.
项数=这个数列中有几个数,即8.
项数很长的时候,可用公式:项数=(末项-首项)÷公差+1 公差即2个连着的数的差.
(2)根据题意,把0.573×25变成57.3×0.25,由乘法分配律,逐步解答即可.
点评:本题考查的目的是根据数字特点经过“转化”变形,巧用等查数列求和、乘法分配律进行简便计算即可.
=(1993+2000)×8÷2,
=3993×4,
=15972.
(2)(0.75×42.7+57.3-0.573×25)÷3×7972,
=(0.75×42.7+57.3-57.3×0.25)÷3×7972,
=[0.75×42.7+57.3×(1-0.25)]÷3×7972,
=[0.75×42.7+57.3×0.75]÷3×7972,
=[0.75×(42.7+57.3)]÷3×7972,
=0.75×100÷3×7972,
=75÷3×7972,
=25×7972,
=25×(4×1993),
=25×4×1993,
=100×1993,
=199300.
分析:(1)这是一道典型的等差数列求和.首项与末项之和×项数÷2,
首项=这个数列的第一个数1993.末项=最后一个数2000.
项数=这个数列中有几个数,即8.
项数很长的时候,可用公式:项数=(末项-首项)÷公差+1 公差即2个连着的数的差.
(2)根据题意,把0.573×25变成57.3×0.25,由乘法分配律,逐步解答即可.
点评:本题考查的目的是根据数字特点经过“转化”变形,巧用等查数列求和、乘法分配律进行简便计算即可.
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目