求和Sn=1/1*4+1/4*7+.1/(3n-2)(3n+1)
问题描述：
求和Sn=1/1*4+1/4*7+.1/(3n-2)(3n+1)
答
裂项求和
1/1*4=1/3*(1/1-1/4)
1/4*7=1/3*(1/4-1/7)
……
1/(3n-2)(3n+1)=1/3*[1/(3n-2)-1/(3n+1)]
把所有的式子加起来，中间的项都可以消去，就得到结果了。思路是这样，自己试一下
答
sn=1/3{(1-1/4)+(1/4-1/7)+...+[1/(3n-2)-1/(3n+1)] }=1/3{1-1/(3n+1)}=n/(3n+1)
答
答：Sn=1/1*4+1/4*7+..+1/(3n-2)(3n+1)=(1-1/4)*1/3+(1/4-1/7)*1/3+...+(1/(3n-2)-1/(3n+1))*1/3=(1-1/4+1/4-1/7+...+1/(3n-2)-1/(3n+1))*1/3=(1-1/(3n+1))*1/3=1/3-1/(9n+3)