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

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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
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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
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

这叫裂项求和法

基本裂项式是


如有不懂请追问

望采纳

1/(3n-1)-1/(3n+2)
通分=3/(3n-1)(3n+2)

23X5/221/2-5/6X1/2(5/12-7/18)X363/4X8/9-1/37/9+7/9X85/8X7-5/8X3 1.（-5）+21x8/2-6-592.（-2）/9-7/9-563.（-2）x8-8x1/2+8/1/84.(-9)x5/6+5/65.1/2+3+5/6-7/12 数列1/2X5,1/5X8,1/8X11,、、、1/[（3N-1）X（3N+2）]的前N项和是 1X1+2X2+3X3+4X4+5X5+……+NXN求此数列的S和 1/2x5+1/5x8+1/8x11+.+1/2006x2009+2009x2013用裂项法计算 裂项法1/2X5+1/5x8+1/8x11.1/2009x2012=? 1/2x5 +1/5x8 +1/8x11+……+1/2009x2012=? 数学matlab写代码,1式：3*x1+3.5*x2+4*x3+4.5*x4+5*x5+5.5*x6+6*x7+6.5*x8=89x1+ x2+x3+x4+x5+x6+x7+x8=200≤x1≤43 0≤x2≤59 0≤x3≤39 0≤x4≤41 0≤x5≤27 0≤x6≤28 0≤x7≤34 0≤x8≤21Yn= （x1,x2,x3,x4,x5,x6,x7,x8）求各种满足 画出求1X3+ 2X4+ 3X5+ 4X6+ 5X7+ 6X8+ 7X8的值的程序框图 数学建模的LINGO使用问题modelmin=x0+x11+x8+x9;x0+x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11=11;x0+x1+x3+x4>=4;x1+x2+x4+x5>=4;x2+x3+x5+x6>=4;x3+x4+x6+x7>=3;x4+x5+x7+x8>=3;x5+x6+x8+x9>=4;x6+x7+x9+x10>=4;x7+x8+x10+x11>=2;x8+x9+x11+x0>=2;x9+x10+x0+x1>=1; x1-x2+x3=1 x2-x3+x4=2 x3-x4+x5=3 x4-x5+x1=4 x5-x1+x2=5 求x1,x2,x3,x4,x5 线性问题在lingo中的表示minZ=5*x1+7*x2+5*x3+4*x4+6*x5+5*x6+5*x7+3*x8 限制条件25*x1+36*x2+32*x3+15*x4+31*x5+28*x6+22*x7+12*x8>=105.69;然后x1到x8必须是1或者0.这个怎么在lingo软件中正确表示出来 matlab ga函数怎么处理有连续和离散混合变量的问题比如：X1、X2、X3、X4、X5为连续变量,范围为[-10,10]；X6、X7、X8、X9、X10为离散变量,可以取的值有：{-8,-5,-2,1,4,7,10}；等式约束条件有：X1+X3+X5+X7 1/2x5+1/5x8+1/8x11+1/11x14+1/14x17+1/17x20 2x5/1+5x8/1+8x11/1+.2006x2009/1+2009x2012/1 用裂项法计算 1/2x5+1/5x8+1/8x11+.+1/32x35要有 详细的算式方法 1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9+9x10+10x11 1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9=