题目

1／（3+6）+1／（3+6+9）+1／（3+6+9+12）+．．．+1／（3+6+9+．．．+150）＝?

提问时间：2020-08-02

答案

1/3+1/(3+6)+1/(3+6+9)+...+1/(3+6+9+...+150)
=1/3+1/[3*(1+2)]+1/[3*(1+2+3)]+...+1/[3*(1+2+3+...+50)]
=2/3*[1/(1*2)+1/(2*3)+1/(3*4)+...+1/(50*51)]
=2/3*[1-1/2+1/2-1/3+1/3-1/4+...+1/50-1/51]
=2/3*[1-1/51]
=2/3*50/51
=100/153