bn=a1+2a2+3a3+4a4+……+nan若an是等差数列,则bn=?

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/20 15:16:24

数列｛an｝是正项等差数列,若bn＝(a1+2a2+3a3+…+nan)/(1＋2＋3＋…+n),则数列｛bn｝也为等差数列
设an公差为d,则
bn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n)
=2(a1+2a2+3a3+…+nan)/n(n+1)
=2(a1+2(a1+d)+3(a1+2d)+…+n(a1+(n-1)d)/n(n+1)
=2{(a1+2a1+3a1+…+na1)+[1*2+2*3+3*4+…(n-1)n]d}/n(n+1)
=2{(n(n+1)a1/2)+[1*2+2*3+3*4+…(n-1)n]d}/n(n+1)
={(n(n+1)a1)+2[1*2+2*3+3*4+…(n-1)n]d}/n(n+1)
=a1+2[1*2+2*3+3*4+…+(n-1)n]d/n(n+1)
=a1+2[1+2+3+…+n-1+1^2+2^2+3^2+…+(n-1)^2]d/n(n+1)
=a1+2(n-1)n(n+1)d/3n(n+1)
=a1+(n-1)2d/3
即是bn是以a1为首数,2d/3为公差的等差数列,证毕.
bn＝a1+2a2+3a3+…nan/1+2+3…+n
b(n+1)=[a1+2a2+3a3+…nan+(n+1)a(n+1)]/[1+2+3…+n+(n+1)]
[n(n+1)/2]bn=a1+2a2+3a3+…nan ①
[(n+1)(n+2)/2]b(n+1)=a1+2a2+3a3+…nan+(n+1)a(n+1) ②
②-①得
[(n+1)(n+2)/2]b(n+1)-[n(n+1)/2]bn=（n+1)a(n+1)
两边同时消去（n+1)得
a(n+1)=[(n+2)/2]b(n+1）-(n/2)bn③
an=[(n+1)/2]bn-[(n-1)/2]b(n-1) ④
③-④得a(n+1)-an=[(n+1)/2]b(n+1）+1/2b(n+1)-[(n+1）/2]bn-[(n-1)/2]bn+[(n-1)/2]b(n-1)-1/2bn
=[(n+1)/2][b(n+1)-bn]+1/2[b(n+1）-bn]-[(n-1)/2][bn-b(n-1)]
又{bn}为等差数列,设公差为d
则a(n+1)-an=[(n+1)/2]d+1/2*d-[(n-1)/2]d
=3/2d
所以{an}是公差为3/2d的等差数列
注：此中的an,bn,a(n+1),b(n+1)均是数列中的项

bn=(a1+2a2+3a3+4a4+……+nan)/(1+2+3+4+……+n)证明an是等差数列是bn是等差数列的充一道数学数列题设两个数列{An},{Bn}满足Bn=（A1+A2+A3+……+nAn)/(1+2+3+……+),若{Bn 设数列an,bn满足:bn=(a1+a2+a3+a4+...+an)/n,若bn是等差数列,求证an也是等差数列 bn}是首项为1,公差4/3的等差数列,且bn=（a1+2a2+……+nan）/（1+2+……+n）, 1.求证{an} 已知:bn=(a1+2a2+...+nan)/(1+2+...+n),数列an成等差数列的充要条件是bn也是等差数列. 已知数列{an}和{bn}满足关系:bn=(a1+a2+a3+…+an)/n,(n∈N*).若{bn}是等差数列,求证{ 已知等差数列an中,a1+a3+a5=21,a2+a4+a6=27,Sn=4Sn=3bn-a1 求an,bn 若数列An是等差数列,则有数列Bn=a1+a2+a3+a4+...+an/n也是等差数列,类比上述性质,相应的,若数列C an是等差数列,bn 是等比数列,a1+b1=3,a2+b2=7,a3+b3=15,a4+b4=35,求an+bn=? 归纳推理：an为等差数列且bn=(a1+2a2+...+nan)/(1+2+3+...+n) 则bn为等差数列那么cn为有两个等差数列{an],{bn]满足（a1+a2+a3+…an）/（b1+b2+b3+…bn）=（7n+2）/（n+3）设an是等差数列,求证以bn=(a1+a2+a3+…+an)/n,n属于N+为通项公式的数列bn是等差数列