首先用等差求和公式求出分母来，就是(2+n)*n/2,然后每一项就是2/(1+n)*n,此项可以写成2/n - 2/(1+n),那么把每一项都分开就只剩下第一项和最后一项，结果就是2-2/101=200/101
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.....+1/(1+2+3+....+100)
=2/(1*2)+2/(2*3)+2/(3*4)+……+2/(100*101)
=2[(1-1/2)+(1/2-1/3)+……+(1/100-1/101)
=2*(1-1/101)
=200/101
1+1/(1+2)+……+1/(1+2+3+……+100)
=1+2/(2*3)+2/(3*4)+……+2/(100*101)
=1+2[1/(2*3)+……+1/(100*101)]
=1+2*[(1/2-1/3)+(1/3-1/4)+……+(1/100-1/101)]
=1+2*(1/2-1/101)
=1+99/101
=200/101
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