n+n+1+n+2+分之一裂项
@浦爸4247：级数[n(n+1)(n+2)]分之一的和 -
奚厚13280582003…… 1/[n(n+1)(n+2)]=1/2*【1/[(n(n+1)]--1/[(n+1)(n+2)]】 =1/2*【1/n--1/(n+1)】--1/2*【1/(n+1)--1/(n+2)】 因此级数的前n项的和为 1/2*【1--1/(n+1)】--1/2*【1/2--1/(n+2)】 当n趋于无穷时,和趋于1/2*【1--1/2】=1/4.
@浦爸4247：级数[n(n+1)(n+2)]分之一的和 - 作业帮
奚厚13280582003…… [答案] 1/[n(n+1)(n+2)]=1/2*【1/[(n(n+1)]--1/[(n+1)(n+2)]】 =1/2*【1/n--1/(n+1)】--1/2*【1/(n+1)--1/(n+2)】 因此级数的前n项的和为 1/2*【1--1/(n+1)】--1/2*【1/2--1/(n+2)】 当n趋于无穷时,和趋于1/2*【1--1/2】=1/4.
@浦爸4247：(n+1)(n+2)(n+3)分之一如何裂项,过程.
奚厚13280582003…… 1/(n+1)(n+2)(n+3)=(1/2)[1/(n+1)(n+2) -1/(n+2)(n+3)]
@浦爸4247：裂项法求和 n乘(n+2) 分之一 -
奚厚13280582003…… 1/n(n+2)=[1/n-1/(n+2)]/2 Sn=[1-1/3+1/2-1/4+1/3-1/5+...+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]/2 =[1+1/2-1/(n+1)-1/(n+2)]/2 =[3/2-(n+2+n+1)/(n+2)(n+1)]/2 =(3n^2+5n)/2(n^2+3n+2) n^2表示n的平方
@浦爸4247：证明:n(n+1)(n+2)分之一+n+1分之一=n(n+2)分之n+1 -
奚厚13280582003…… 左边通分得: 左边 = 1/{n(n+1)(n+2)} + {n(n+2)}/{n(n+1)(n+2)} = {1+n(n+2)}/{n(n+1)(n+2)} = {1+n²+2n}/{n(n+1)(n+2)} = (n+1)²/{n(n+1)(n+2)} = (n+1)/{n(n+2)} = 右边
@浦爸4247：n(n+1)分之1+(n+1)(n+2)分之1 -
奚厚13280582003…… n(n+1)分之1+(n+1)(n+2)分之1+(n+2)(n+3)分之1+……+(n+9)(n+10)分之1、 =[1/n-1/(n+1)]+[1/(n+1)-1/(n+2)]+[1/(n+2)-1/(n+3)]+.....+[1/(n+9)-1/(n+10)] =1/n-1/(n+10) =10/[n(n+10)]
@浦爸4247：n+1分之一加n+2分之一一直加到2n分之一 怎样求最小值? -
奚厚13280582003…… 1/(n+1)+1/(n+2) = 1/(2n) 2n*[1/(n+1)+1/(n+2)] = 2n*[1/(2n)] 2n/(n+1) + 2n/(n+2) = 1 (n+1)(n+2)*[2n/(n+1) + 2n/(n+2)] = 1*(n+1)(n+2) 2n*(n+2) + 2n*(n+1) = (n+1)(n+2) 2n^2 + 4n + 2n^2 + 2n= n^2 + 2n + n + 2 4n^2 + 6n = n^2 + 3n + 2 3n^2 + 3n - 2 ...
@浦爸4247：an=(n+1)(n+2)分之一,求sn -
奚厚13280582003…… an=1/[(n+1)*(n+2)]=[1/(n+1)]-[1/(n+2)] 则:Sn=[1/2-1/3]+[1/3-1/4]+[1/5-1/6]+…+[1/(n+1)-1/(n+2)] Sn=1/2-1/(n+2) Sn=n/(2n+4)
@浦爸4247：裂项相消法:1/[n(n+1)(n+2)]=1/2*{1/[n(n+1)] - 1/[(n+1)(n+2)}如何理解?急用,高手赐教!
奚厚13280582003…… 1/[n(n+1)(n+2)]=(1/2)(n+2-n)/[n(n+1)(n+2)] =(1/2){(n+2)/[n(n+1)(n+2)-n/[n(n+1)(n+2)]} =(1/2){1/[n(n+1)]-1/[(n+1)(n+2)]}
@浦爸4247：{n(n+1)(n+2)}分之一怎样求和 - 作业帮
奚厚13280582003…… [答案] 1/(n(n+1)(n+2)=(1/2)*2/n(n+1)(n+2)=(1/2)*[(n+2)-n]/n(n+1)(n+2)=(1/2)*[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]=(1/2)*[1/n(n+1)-1/(n+1)(n+2)]所以1/1*2*3+……+1/n(n+1)(n+2)=(1/2)*[1/1*2-1/2*3+……+1/n(n+1)-1/(n...
奚厚13280582003…… 1/[n(n+1)(n+2)]=1/2*【1/[(n(n+1)]--1/[(n+1)(n+2)]】 =1/2*【1/n--1/(n+1)】--1/2*【1/(n+1)--1/(n+2)】 因此级数的前n项的和为 1/2*【1--1/(n+1)】--1/2*【1/2--1/(n+2)】 当n趋于无穷时,和趋于1/2*【1--1/2】=1/4.
@浦爸4247：级数[n(n+1)(n+2)]分之一的和 - 作业帮
奚厚13280582003…… [答案] 1/[n(n+1)(n+2)]=1/2*【1/[(n(n+1)]--1/[(n+1)(n+2)]】 =1/2*【1/n--1/(n+1)】--1/2*【1/(n+1)--1/(n+2)】 因此级数的前n项的和为 1/2*【1--1/(n+1)】--1/2*【1/2--1/(n+2)】 当n趋于无穷时,和趋于1/2*【1--1/2】=1/4.
@浦爸4247：(n+1)(n+2)(n+3)分之一如何裂项,过程.
奚厚13280582003…… 1/(n+1)(n+2)(n+3)=(1/2)[1/(n+1)(n+2) -1/(n+2)(n+3)]
@浦爸4247：裂项法求和 n乘(n+2) 分之一 -
奚厚13280582003…… 1/n(n+2)=[1/n-1/(n+2)]/2 Sn=[1-1/3+1/2-1/4+1/3-1/5+...+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]/2 =[1+1/2-1/(n+1)-1/(n+2)]/2 =[3/2-(n+2+n+1)/(n+2)(n+1)]/2 =(3n^2+5n)/2(n^2+3n+2) n^2表示n的平方
@浦爸4247：证明:n(n+1)(n+2)分之一+n+1分之一=n(n+2)分之n+1 -
奚厚13280582003…… 左边通分得: 左边 = 1/{n(n+1)(n+2)} + {n(n+2)}/{n(n+1)(n+2)} = {1+n(n+2)}/{n(n+1)(n+2)} = {1+n²+2n}/{n(n+1)(n+2)} = (n+1)²/{n(n+1)(n+2)} = (n+1)/{n(n+2)} = 右边
@浦爸4247：n(n+1)分之1+(n+1)(n+2)分之1 -
奚厚13280582003…… n(n+1)分之1+(n+1)(n+2)分之1+(n+2)(n+3)分之1+……+(n+9)(n+10)分之1、 =[1/n-1/(n+1)]+[1/(n+1)-1/(n+2)]+[1/(n+2)-1/(n+3)]+.....+[1/(n+9)-1/(n+10)] =1/n-1/(n+10) =10/[n(n+10)]
@浦爸4247：n+1分之一加n+2分之一一直加到2n分之一 怎样求最小值? -
奚厚13280582003…… 1/(n+1)+1/(n+2) = 1/(2n) 2n*[1/(n+1)+1/(n+2)] = 2n*[1/(2n)] 2n/(n+1) + 2n/(n+2) = 1 (n+1)(n+2)*[2n/(n+1) + 2n/(n+2)] = 1*(n+1)(n+2) 2n*(n+2) + 2n*(n+1) = (n+1)(n+2) 2n^2 + 4n + 2n^2 + 2n= n^2 + 2n + n + 2 4n^2 + 6n = n^2 + 3n + 2 3n^2 + 3n - 2 ...
@浦爸4247：an=(n+1)(n+2)分之一,求sn -
奚厚13280582003…… an=1/[(n+1)*(n+2)]=[1/(n+1)]-[1/(n+2)] 则:Sn=[1/2-1/3]+[1/3-1/4]+[1/5-1/6]+…+[1/(n+1)-1/(n+2)] Sn=1/2-1/(n+2) Sn=n/(2n+4)
@浦爸4247：裂项相消法:1/[n(n+1)(n+2)]=1/2*{1/[n(n+1)] - 1/[(n+1)(n+2)}如何理解?急用,高手赐教!
奚厚13280582003…… 1/[n(n+1)(n+2)]=(1/2)(n+2-n)/[n(n+1)(n+2)] =(1/2){(n+2)/[n(n+1)(n+2)-n/[n(n+1)(n+2)]} =(1/2){1/[n(n+1)]-1/[(n+1)(n+2)]}
@浦爸4247：{n(n+1)(n+2)}分之一怎样求和 - 作业帮
奚厚13280582003…… [答案] 1/(n(n+1)(n+2)=(1/2)*2/n(n+1)(n+2)=(1/2)*[(n+2)-n]/n(n+1)(n+2)=(1/2)*[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]=(1/2)*[1/n(n+1)-1/(n+1)(n+2)]所以1/1*2*3+……+1/n(n+1)(n+2)=(1/2)*[1/1*2-1/2*3+……+1/n(n+1)-1/(n...
