(1+1/1+2)+(1/1+2+3)+...+(1/1+2+3+...+100)

来源：百度知道编辑：UC知道时间：2024/05/15 14:03:42

急用

先从分母找规律:1+2=(2*3)/2,1+2+3=(3*4)/2,1+2+3+4=(4*5)/2...1+2+3+...+100=(100*101)/2;每项可分解为:1/(1+2)=1/[(2*3)/2]=2/(2*3)=2(1/2-1/3),1/(1+2+3)=1/[(3*4)/2]=2/(3*4)=2(1/3-1/4)...1/(1+2+3+...+100)=1/[(100*101)/2]=2/(100*101)=2(1/100-1/101);所以原式=1+2(1/2-1/3)+2(1/3-1/4)+2(1/4-1/5)...+2(1/100-1/101)=1+2*(1/2+1/3-1/3+1/4-1/4+...1/100-1/101)=1+2(1/2-1/101)=2-2/101=200/101.

1000

(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1) (1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000) 1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100) 1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100) 1+1/1+2+1/1+2+3+...+1/1+2+3...+2000 1+1/1+2+1/1+2+3.........+1/1+2+3.....100 1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=? 1+1/2+1/3+.....+1/n 1+1/2+1/3+...+1/100 1-1/2+1/3-.....-1/10