1+1/(1+2)+1/(1+2+3)+···1/(1+2+3+···+100)等于几？

来源：百度知道编辑：UC知道时间：2024/06/21 10:29:12

1+2+3+……+n=n(n+1)/2
1/(1+2+3+……+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
所以1+1/(1+2)+1/(1+2+3)+···1/(1+2+3+···+100)
=1+2*[(1/2-1/3)+(1/3-1/4)+……+(1/100-1/101)]
=1+2*(1/2-1/101)
=2-2/101
=200/101

1.
1+2+3+……+n=n(n+1)/2
1/(1+2+3+……+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
所以1+1/(1+2)+1/(1+2+3)+···1/(1+2+3+···+100)
=1+2*[(1/2-1/3)+(1/3-1/4)+……+(1/100-1/101)]
=1+2*(1/2-1/101)
=2-2/101
=200/101
2.
通项为1/(1+2+…+n)=1/[(n+1)n/2]=2/[n(n+1)]
=2[1/n-1/(n+1)]
所以
1+1/(1+2)+……+1/(1+2+…+n)
=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2[1-1/(n+1)];
代入n=100得到
2[1-1/101]=200/101

通项为1/(1+2+…+n)=1/[(n+1)n/2]=2/[n(n+1)]
=2[1/n-1/(n+1)]
所以
1+1/(1+2)+……+1/(1+2+…+n)
=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2[1-1/(n+1)];
代入n=100得到
2[1-1/101]=200/101

1/(1+1)*1/2+1/(1+2)*2/2+1/(1+3)*3/2+……+1/(1+n)n/2
=1/2[1/1*2+1/2*3+1/3*4+1/n*(n+1)]
=1/2[1/1-1/2+1/2-1/3+1/3-1/4+……-1/

(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)