用通分算法知道：1/[n(n+1)] = 1/n - 1/(n+1)
所以：
1*2分之1+2*3分之1+3*4分之1+……+199*200分之1
=(1-1/2) + (1/2-1/3) + (1/3-1/4) + ...+ (1/199-1/200)
=1 + (-1/2+1/2) + (-1/3+1/3)...-1/200
=1-1/200
=199/200