數(shù)列｛an｝是正項(xiàng)等差數(shù)列,若bn＝(a1+2a2+3a3+…+nan)/(1＋2＋3＋…+n),則數(shù)列｛bn｝也為 等差數(shù)列
設(shè)an公差為d,則
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為首數(shù),2d/3為公差的等差數(shù)列,證畢.
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)
兩邊同時(shí)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}為等差數(shù)列,設(shè)公差為d
則a(n+1)-an=[(n+1)/2]d+1/2*d-[(n-1)/2]d
=3/2d
所以{an}是公差為3/2d的等差數(shù)列
注：此中的an,bn,a(n+1),b(n+1)均是數(shù)列中的項(xiàng)