答：题目应该缺少了大量的括号吧?
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+100)
第n项的分母是自然数之和(n+1)*n/2
所以：
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+100)
=1+2/(2*3)+2/(3*4)+2/(4*5)+...+2/(100*101)
=1+2*(1/2-1/3+1/3-1/4+1/4-1/5+...+1/100-1/101)
=1+2*(1/2-1/101)
=1+1-2/101
=2-2/101
=200/101