3+6+9+…++300+3031999+999×9991.24×0.25+124×750.8888×6+0.1111×52.

发布时间:2020-07-31 05:35:52

3+6+9+…++300+3031999+999×9991.24×0.25+124×750.8888×6+0.1111×52.

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解:(1)3+6+9+…++300+303
=(3+303)×[(303-3)÷3+1]÷2,
=306×[300÷3+1]÷2,
=306×101÷2,
=15453;

(2)1999+999×999
=1000+999+999×999,
=1000+(999+1)×999,
=1000+1000×999,
=(999+1)×1000,
=1000×1000,
=1000000.

(3)1.24×0.25+124×75
=0.31×(4×0.25)+31×(4×25)×3,
=0.31×1+31×3×100,
=0.31+9300,
=9300.31;

(4)0.8888×6+0.1111×52
=0.1111×(8×6)+0.1111×52,
=0.1111×48+0.1111×52,
=0.1111×(48+52),
=0.1111×100,
=11.11.
解析分析:(1)可根据高斯求和公式有关公式进行计算:项数=(尾项-首项)÷公差+1,等差数列和=(首项+尾项)×项数÷2;(2)可将1999拆分成1000+999后根据乘法分配律计算;(3)可将1.24×0.25变为0.31×4×0.25,将124×75变为31×4×25×3后进行巧算;(4)可将0.8888×6变为0.1111×8×6后根据乘法分配律巧算.

点评:完成此类题目要细心分析式中数据的特点及内在联系,然后运用合适的方法时行计算.
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