急·····················1+2+3+...+999=?

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/27 17:52:29
急·····················1+2+3+...+999=?

急·····················1+2+3+...+999=?
急·····················
1+2+3+...+999=?

急·····················1+2+3+...+999=?
这个是等差数列用等差数列求和公式:(末项+首项)*项数/2和等差数列求项数公式:(末项-首项)/公差+1(999-1)/1+1=999(1+999)*999/2=499500

(1+999)*999/2=499500

1+2+3+...+999=(1+999)*999/2=499500

等差数列求和公式n*(n+1)/2
结果为499500

(1+999)×999除2=