分类：教育/科学-理工学科-数学
问题：谁能解数学题1+（1/1+2）+(1/1+2+3)+...... 
1+（1/1+2）+(1/1+2+3)+........+(1/1+2+3+....n)=
回答：通项为1/[n(n+1)/2]=2×[(1/n)-1/(n+1)]

1+[1/(1+2)]+[1/(1+2+3)]+……+[1/(1+2+3+....n)]
=2×[1-(1/2)]+2×[(1/2)-(1/3)]+2×[(1/3)-(1/4)]+……+2×[(1/99)-(1/100)]+2×[(1/100)-(1/101)]
=2×[1-(1/2)+(1/2)-(1/3)+(1/3)+……+(1/99)-(1/100)+(1/100)-(1/101)]
=2×(1-1/101)
=200/101