Question

计算1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……1/(1+2+3+……+100)

Answer 1

注意观察，第 N 个加式可以表述成：

1/(1 + 2 + 3 + ... + n)
= 1/[n(n + 1)/2]
= 2/[n(n + 1)]
= 2[1/n - 1/(n + 1)]

那么有：

1/1 + 1/(1 + 2) + 1/(1 + 2 + 3) + .... + 1/(1 + 2 + 3 + ... + 100)
= 1 + 2/(2*3) + 2/(3*4) + .... + 2/(100*101)
= 1 + 2*(1/2 - 1/3) + 2*(1/3 - 1/4) + ... + 2*(1/100 - 1/101)
= 1 + 2*(1/2 - 1/101)
= 2 - 2/101
= 200/101

Answer 2

1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……1/(1+2+3+……+100)
=1+1/3+1/6+1/10+-----+1/5050
=2×[1/2+1/6+1/12+1/20+------+1/(101×100)]
=2×(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+------+1/100-1/101)
=2×(1-1/101)
=200/101