数列1/2X5,1/5X8,1/8X11,、、、1/[（3N-1）X（3N+2）]的前N项和是

1/2X5+1/5X8+1/8X11+...+1/[（3N-1）X（3N+2）]
=1/3(1/2-1/5+1/5-1/8+...+1/(3n-1)-1/(3n+2))
=1/3*(1/2-1/(3n+2))
=n/[2(3n+2)]1/3*(1/2-1/(3n+2))这一步怎么算的，请解中间的项消去了啊，只剩第一项和最后一项能把过程也做一下吗我已经把过程写全了啊这个的解啊，我还是不明白1/3*(1/2-1/(3n+2))1/2-1/5+1/5-1/8+...+1/(3n-1)-1/(3n+2)看到了吗？相邻的项的和是0即＝1/2+（-1/5+1/5）+（-1/8+...+1/(3n-1)-1/(3n+2)看明白了吗？这懂啦，是最后的第二步的解不懂为什么是=n/[2(3n+2)]这个是通分呀1/3*(1/2-1/(3n+2))=1/3*[(3n+2)-2]/[2(3n+2)]=n/[2(3n+2)]