bn=(a1+2a2+...+nan)/(1+2+...+n)
a1+2a2+...+nan=(1+2+...+n)bn=n(n+1)bn/2 (1)
a1+2a2+...(n-1)an=n(n-1)b(n-1)/2 (2)
(1)-(2)
nan=n(n+1)bn/2 -n(n-1)b(n-1)/2
an=(n+1)bn/2 -(n-1)b(n-1)/2
a(n+1)=(n+2)b(n+1)/2-nbn/2
1.
數(shù)列{an}是等差數(shù)列時(shí),a(n+1)-an為定值.
(n+2)b(n+1)/2 -nbn/2 -(n+1)bn/2 +(n-1)b(n-1)/2
=[[b(n+1)-2bn+b(n-1)/2]n +[2b(n+1)-bn-b(n-1)]/2
要對(duì)任意正整數(shù)n,[[b(n+1)-2bn+b(n-1)/2]n +[2b(n+1)-bn-b(n-1)]/2恒為定值,則只有n項(xiàng)系數(shù)=0
[b(n+1)-2bn+b(n-1)]/2=0
b(n+1)-2bn+b(n-1)=0
b(n+1)-bn=bn-b(n-1)
數(shù)列{bn}是等差數(shù)列.
2.
數(shù)列{bn}是等差數(shù)列時(shí),設(shè)公差為d
an=(n+1)bn/2 -(n-1)b(n-1)/2=(n+1)bn/2 -(n-1)(bn -d)/2=bn+ (n-1)d/2
a(n+1)=(n+2)b(n+1)/2-nbn/2=(n+2)(bn +d)/2 -nbn/2=bn +(n+2)d/2
a(n+1)-an=bn+(n+2)d/2 -bn -(n-1)d/2=(3/2)d,為定值.
數(shù)列{an}是等差數(shù)列.
綜上,得數(shù)列an成等差數(shù)列的充要條件是bn也是等差數(shù)列.