一张纸,其厚度为a ,面积为b,先将此张纸对折n次求和Sn=1/1*4+1/4*7+...+1/(3n-2)*(3n+1)=?
问题描述：
一张纸,其厚度为a ,面积为b,先将此张纸对折n次
求和Sn=1/1*4+1/4*7+...+1/(3n-2)*(3n+1)=?
答
因为1/(3n-2)*(3n+1)=[1/(3n-2）-1/(3n+1）]/3
所以Sn=1/3[(1/1-1/4)+(1/4-1/7)+..........1/(3n+1）]=n/(3n+1)
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答
1/(3n-2)*(3n+1)=3*[1/(3n-2)-1/(3n+1)]
Sn=1/1*4+1/4*7+...+1/(3n-2)*(3n+1)
=3*[1/1-1/4+1/4-1/7+.....+1/(3n-2)-1/(3n+1)]
=3*[1-1/(3n+1)]
=9*n/(3n+1)
答
1/(3n-2)*(3n+1)=1/3*[1-1/(3n+1)]
Sn= 1/3*[1/1-1/4+1/4-1/7+.+1/(3n-2)-1/(3n+1)]=n/(3n+1)