您的位置: 首页  >  作业答案  >  其他  >  求数列1/2x5,1/5x8,1/8x11...,1/(3n-1)(3n+2)...的前n项和=1/3[1/(3n-1)-1/(3n+2)]=1/3(1/2-1/5)+1/3(1/5-1/8)+1/3……1/3[1/(3n-1)-1/(3n+2)]=1/3[1/2-1/5+1/5-1/8+1/8-……+1/(3n-1)-1/(3n+2)].=n/[2(3n+2)]=1/3[1/(3n-1)-1/(3n+2)]这步是如何得到的?

求数列1/2x5,1/5x8,1/8x11...,1/(3n-1)(3n+2)...的前n项和=1/3[1/(3n-1)-1/(3n+2)]=1/3(1/2-1/5)+1/3(1/5-1/8)+1/3……1/3[1/(3n-1)-1/(3n+2)]=1/3[1/2-1/5+1/5-1/8+1/8-……+1/(3n-1)-1/(3n+2)].=n/[2(3n+2)]=1/3[1/(3n-1)-1/(3n+2)]这步是如何得到的?

分类: 作业答案 • 2021-12-19 17:15:14
问题描述：

求数列1/2x5,1/5x8,1/8x11...,1/(3n-1)(3n+2)...的前n项和
=1/3[1/(3n-1)-1/(3n+2)]
=1/3(1/2-1/5)+1/3(1/5-1/8)+1/3……1/3[1/(3n-1)-1/(3n+2)]
=1/3[1/2-1/5+1/5-1/8+1/8-……+1/(3n-1)-1/(3n+2)]
.
=n/[2(3n+2)]
=1/3[1/(3n-1)-1/(3n+2)]这步是如何得到的?

答

这叫裂项求和法

基本裂项式是


如有不懂请追问

望采纳

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