1-2/1×(1+2)-3/(1+2)×(1+2+3)1.1-2/1×(1+2)-3/(1+2)×(1+2+3)-4/(1+2+3)×(1+2+3+4)-...-10/(1+2+...+9)×(1+2+...+10)怕大家看不懂,用文字形式说明.1减去1×(1+2)分之2,减去(1+2)×(1+2+3)分之3,减去(1+2+3)×(1+2+3+4)分之4...以此类
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 06:07:03
![1-2/1×(1+2)-3/(1+2)×(1+2+3)1.1-2/1×(1+2)-3/(1+2)×(1+2+3)-4/(1+2+3)×(1+2+3+4)-...-10/(1+2+...+9)×(1+2+...+10)怕大家看不懂,用文字形式说明.1减去1×(1+2)分之2,减去(1+2)×(1+2+3)分之3,减去(1+2+3)×(1+2+3+4)分之4...以此类](/uploads/image/z/12667014-54-4.jpg?t=1-2%2F1%C3%97%281%2B2%29-3%2F%281%2B2%29%C3%97%281%2B2%2B3%291.1-2%2F1%C3%97%281%2B2%29-3%2F%281%2B2%29%C3%97%281%2B2%2B3%29-4%2F%281%2B2%2B3%29%C3%97%281%2B2%2B3%2B4%29-...-10%2F%281%2B2%2B...%2B9%29%C3%97%281%2B2%2B...%2B10%29%E6%80%95%E5%A4%A7%E5%AE%B6%E7%9C%8B%E4%B8%8D%E6%87%82%2C%E7%94%A8%E6%96%87%E5%AD%97%E5%BD%A2%E5%BC%8F%E8%AF%B4%E6%98%8E.1%E5%87%8F%E5%8E%BB1%C3%97%281%2B2%29%E5%88%86%E4%B9%8B2%2C%E5%87%8F%E5%8E%BB%281%2B2%29%C3%97%281%2B2%2B3%29%E5%88%86%E4%B9%8B3%2C%E5%87%8F%E5%8E%BB%281%2B2%2B3%29%C3%97%281%2B2%2B3%2B4%29%E5%88%86%E4%B9%8B4...%E4%BB%A5%E6%AD%A4%E7%B1%BB)
1-2/1×(1+2)-3/(1+2)×(1+2+3)1.1-2/1×(1+2)-3/(1+2)×(1+2+3)-4/(1+2+3)×(1+2+3+4)-...-10/(1+2+...+9)×(1+2+...+10)怕大家看不懂,用文字形式说明.1减去1×(1+2)分之2,减去(1+2)×(1+2+3)分之3,减去(1+2+3)×(1+2+3+4)分之4...以此类
1-2/1×(1+2)-3/(1+2)×(1+2+3)
1.1-2/1×(1+2)-3/(1+2)×(1+2+3)-4/(1+2+3)×(1+2+3+4)-...-10/(1+2+...+9)×(1+2+...+10)
怕大家看不懂,用文字形式说明.1减去1×(1+2)分之2,减去(1+2)×(1+2+3)分之3,减去(1+2+3)×(1+2+3+4)分之4...以此类推,一直减到(1+2+...+9)×(1+2+...+10)分之10
1-2/1×(1+2)-3/(1+2)×(1+2+3)1.1-2/1×(1+2)-3/(1+2)×(1+2+3)-4/(1+2+3)×(1+2+3+4)-...-10/(1+2+...+9)×(1+2+...+10)怕大家看不懂,用文字形式说明.1减去1×(1+2)分之2,减去(1+2)×(1+2+3)分之3,减去(1+2+3)×(1+2+3+4)分之4...以此类
是典型的使用裂项求和的题目:
由于n/((1+2+3+...n-1)*(1+2+..n)=1/((1+2+...n-1)-1/(1+2+...n)
从而原式为
1-(1-1/(1+2))-(1/(1+2)-1/(1+2+3))-(1/(1+2+3)-1/(1+2+3+4))-.-(1/(1+2+3..+n-1)-1/(1+2+..+n))=1-1/(1+2+3+..n)=(2+3+..+n)/(1+2+3..+n)
=(n+2)*(n-1)/(n*(n+1))
此题已将数字拆开,就是提示你可以裂项
1+2+3+4+……+n=n*(n+1)/2
以此公式全部化简约分即可