Question

1/2+1/（2+3）+1/（2+3+4)......+1/(2+3+4.......+200) 简便计算

Answer 1

2+3+4+...+n=(n+2)(n-1)/2
所以1/(2+3+4+...+n)=2/[(n-1)(n+2)]=2/3[(n+2)-(n-1)]/[(n+2)(n-1)]=2/3[1/(n-1)-1/(n+2)]
所以1/2+1/（2+3）+1/（2+3+4)......+1/(2+3+4.......+200)
=1/2+2/3(1/2-1/5+1/3-1/6+1/4-1/7+1/5-1/8+...+1/196-1/199+1/197-1/200+1/198-1/201+1/199-1/202)
=1/2+2/3[1/2+1/3+1/4+1/5+...+1/199-(1/5+...+1/199+1/200+1/201+1/202)]
=1/2+2/3(1/2+1/3+1/4-1/200-1/201-1/202)
剩下的就只有硬算了