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：2024/06/01 02:32:00

200/101

1/(1+2+…+n)=2/(n(n+1))=2( 1/n - 1/(n+1) ) 故原式=2（1/1-1/2+1/2-1/3+1/3-1/4……+1/99-1/100+1/100-1/101)=2(1-1/101)=200/101

1/(1+2+…+n)=2/(n(n+1))=2( 1/n - 1/(n+1) )
2（1/1-1/2+1/2-1/3+1/3-1/4……+1/99-1/100+1/100-1/101)=2(1-1/101)=200/101

裂项 1/(1+2+…+n)=2/(n(n+1))=2( 1/n - 1/(n+1) )(补充： 1+2+3+4+....+n=n.(n+1)/2,等差公式求和)
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...100)=
2（1/1-1/2+1/2-1/3+1/3-1/4……+1/99-1/100+1/100-1/101)=2(1-1/101)=200/101

(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1) 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)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000) 1+1/2+1+1/3+1+1/4+......+1/100=? (1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8) (1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)