1+1\(1+2)+1\(1+2+3)+1\(1+2+3+4)+...+1\(1+2+3+...+100)
来源:百度知道 编辑:UC知道 时间:2024/05/26 19:23:31
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首先用等差求和公式求出分母来,就是(2+n)*n/2,然后每一项就是2/(1+n)*n,此项可以写成2/n - 2/(1+n),那么把每一项都分开就只剩下第一项和最后一项,结果就是2-2/101=200/101
在网上查
(1-1\2004)(1-1\2003)(1-1\2002).........(1-1\3)(1-1\2)
(1+1\2)*(1+1\3)*(1+1\3)*(1+1\4)*......(1+1\20)
((1+1\2)(1+1\3)(1+1\4)(1+1\5)------(1+1\100))\(1-1\2)(1-1\3)(1-1\4)------(1-1\100)
((1+1\2)(1+1\3)(1+1\4)(1+1\5)------(1+1\100))\(1-1\2)(1-1\3)(1-1\4)------(1-1\100)等多少
(1\2+1\3+...+1\2006)(1+1\2+1\3+...+1\2005)-(1+1\2+1\3+...+1\2006)(1\2+1\3+...+1\2005)
1+1\(1+2)+1\(1+2+3)+1\(1+2+3+4)+...+1\(1+2+3+...+100)
1\2+1\4+1\8+...+1\256+1\512+1\1024=??
(1-1\2)+(1\2-1\3)+(1\3-1\4)+(1\4-1\5)+(1\5-1\6)+........(1\2005-1\2006
(1-1\2)+(1\2-1\3)+(1\3-1\4)+(1\4-1\5)+(1\5-1\6)+........(1\2005-1\20把括号变为绝对号咋么做06
[1+(-1\2)]+[1\2+(-1\3)]+[1\3+(-1\4)]+......[1\1999+(-1\2000)]