这叫裂项求和法
基本裂项式是
如有不懂请追问
望采纳
求数列1/2x5,1/5x8,1/8x11...,1/(3n-1)(3n+2)...的前n项和
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?
=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?
数学人气：289 ℃时间：2020-02-03 14:51:17
优质解答
我来回答
类似推荐