Question

1+1/1+2+1/1+2+3+1/1+2+3+4+……1/1+2+3+...+100等于多少

Answer 1

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

Answer 2

通项an=2/n(n+1)=2[1/n-1/(n+1)]
∴
1+1/1+2+1/1+2+3+1/1+2+3+4+……1/1+2+3+...+100
=2(1-1/2)+2(1/2-1/3)+……+2(1/100-1/101)
=2（1-1/2+1/2-1/3+……+1/100-1/101）
=200/101

Answer 3

99/50.=1+2*（1/2-1/3）+2*(1/3-1/4)+....+2*(1/99-1/100)

Answer 4