Question

急于向高手求解计算题，请各位能写出解题思路，急！谢谢了

不知怎么录入，只好上传图片了，

Answer 1

等于200/303;
原式=1/3×[1+1/(1+2)+1/(1+2+3)+......+1/(1+2+3+...+100)]
=2/3×(1-1/2+1/2-1/3+1/4-1/4+......+1/100-1/101)
=200/303

Answer 2

观察通项公式an=1/(3+6+9+...+3n)=1/[3(1+2+...+n)]=1/{3*[(1+n)n/2]}=(2/3)*[1/n(n+1)]
=(2/3)*[1/n - 1/(n+1)]
∴原式=(2/3)[1- 1/2] +(2/3)[1/2 - 1/3]+...+(2/3)[1/100 - 1/101]
=(2/3)*(1-1/2+1/2-1/3+...+1/100-1/101)
=2/3*(1-1/101)
=200/303