Question

(1+2分之1）+（1+2+3分之1）+（1+2+3+4分之1）+...+(1+2+...+99分之1)等于多少?

Answer 1

1+2+3+……+n=n(n+1)/2

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

故1/(1+2) +1/(1+2+3)+……+1/(1+2+……99)

=2[1/2-1/3+1/3-1/4+……+1/99-1/100]

=2(1/2 -1/100)

=2(49/100)

=49/50